Mixing
Is there a way to evaluate analytically the following infinite double sum?
$begingroup$
Consider the following double sum
$$
S = sum_{n=1}^infty sum_{m=1}^infty
frac{1}{a (2n-1)^2 - b (2m-1)^2} , ,
$$
where $a$ and $b$ are both positive real numbers given by
begin{align}
a &= frac{1}{2} - frac{sqrt{2}}{32} , , \
b &= frac{1}{4} - frac{3sqrt{2}}{32} , .
end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = frac{pi}{4sqrt{ab}} sum_{m=1}^infty
frac{tan left( frac{pi}{2} sqrt{frac{b}{a}} (2m-1) right)}{2m-1} , .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.
real-analysis calculus sequences-and-series trigonometry summation
asked Jan 15 at 17:20
Math StudentMath Student
130521
$endgroup$
|
show 2 more comments
$begingroup$
Consider the following double sum
$$
S = sum_{n=1}^infty sum_{m=1}^infty
frac{1}{a (2n-1)^2 - b (2m-1)^2} , ,
$$
where $a$ and $b$ are both positive real numbers given by
begin{align}
a &= frac{1}{2} - frac{sqrt{2}}{32} , , \
b &= frac{1}{4} - frac{3sqrt{2}}{32} , .
end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = frac{pi}{4sqrt{ab}} sum_{m=1}^infty
frac{tan left( frac{pi}{2} sqrt{frac{b}{a}} (2m-1) right)}{2m-1} , .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.
real-analysis calculus sequences-and-series trigonometry summation
asked Jan 15 at 17:20
Math StudentMath Student
130521
$endgroup$
2
$begingroup$
Maybe Poisson summation formula?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:28
$begingroup$
@LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated
$endgroup$
– Math Student
Jan 15 at 17:31
4
$begingroup$
On second thoughts, I believe now that the double sum is divergent.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:07
1
$begingroup$
@LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint
$endgroup$
– Math Student
Jan 16 at 7:25
2
$begingroup$
$$ S=frac{pi}{4sqrt{ab}}sum_{m=1}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1} $$
$endgroup$
– Hazem Orabi
Jan 24 at 13:11
|
show 2 more comments
$begingroup$
Consider the following double sum
$$
S = sum_{n=1}^infty sum_{m=1}^infty
frac{1}{a (2n-1)^2 - b (2m-1)^2} , ,
$$
where $a$ and $b$ are both positive real numbers given by
begin{align}
a &= frac{1}{2} - frac{sqrt{2}}{32} , , \
b &= frac{1}{4} - frac{3sqrt{2}}{32} , .
end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = frac{pi}{4sqrt{ab}} sum_{m=1}^infty
frac{tan left( frac{pi}{2} sqrt{frac{b}{a}} (2m-1) right)}{2m-1} , .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.
real-analysis calculus sequences-and-series trigonometry summation
asked Jan 15 at 17:20
Math StudentMath Student
130521
$endgroup$
Consider the following double sum
$$
S = sum_{n=1}^infty sum_{m=1}^infty
frac{1}{a (2n-1)^2 - b (2m-1)^2} , ,
$$
where $a$ and $b$ are both positive real numbers given by
begin{align}
a &= frac{1}{2} - frac{sqrt{2}}{32} , , \
b &= frac{1}{4} - frac{3sqrt{2}}{32} , .
end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = frac{pi}{4sqrt{ab}} sum_{m=1}^infty
frac{tan left( frac{pi}{2} sqrt{frac{b}{a}} (2m-1) right)}{2m-1} , .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.
real-analysis calculus sequences-and-series trigonometry summation
real-analysis calculus sequences-and-series trigonometry summation
asked Jan 15 at 17:20
Math StudentMath Student
130521
asked Jan 15 at 17:20
Math StudentMath Student
130521
edited Jan 24 at 15:20
Math Student
asked Jan 15 at 17:20
Math StudentMath Student
130521
asked Jan 15 at 17:20
Math StudentMath Student
130521
asked Jan 15 at 17:20
Math StudentMath Student
130521
130521
2
$begingroup$
Maybe Poisson summation formula?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:28
$begingroup$
@LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated
$endgroup$
– Math Student
Jan 15 at 17:31
4
$begingroup$
On second thoughts, I believe now that the double sum is divergent.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:07
1
$begingroup$
@LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint
$endgroup$
– Math Student
Jan 16 at 7:25
2
$begingroup$
$$ S=frac{pi}{4sqrt{ab}}sum_{m=1}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1} $$
$endgroup$
– Hazem Orabi
Jan 24 at 13:11
|
show 2 more comments
2
$begingroup$
Maybe Poisson summation formula?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:28
$begingroup$
@LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated
$endgroup$
– Math Student
Jan 15 at 17:31
4
$begingroup$
On second thoughts, I believe now that the double sum is divergent.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:07
1
$begingroup$
@LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint
$endgroup$
– Math Student
Jan 16 at 7:25
2
$begingroup$
$$ S=frac{pi}{4sqrt{ab}}sum_{m=1}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1} $$
$endgroup$
– Hazem Orabi
Jan 24 at 13:11
2
2
$begingroup$
Maybe Poisson summation formula?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:28
$begingroup$
Maybe Poisson summation formula?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:28
$begingroup$
@LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated
$endgroup$
– Math Student
Jan 15 at 17:31
$begingroup$
@LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated
$endgroup$
– Math Student
Jan 15 at 17:31
4
4
$begingroup$
On second thoughts, I believe now that the double sum is divergent.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:07
$begingroup$
On second thoughts, I believe now that the double sum is divergent.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:07
1
1
$begingroup$
@LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint
$endgroup$
– Math Student
Jan 16 at 7:25
$begingroup$
@LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint
$endgroup$
– Math Student
Jan 16 at 7:25
2
2
$begingroup$
$$ S=frac{pi}{4sqrt{ab}}sum_{m=1}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1} $$
$endgroup$
– Hazem Orabi
Jan 24 at 13:11
$begingroup$
$$ S=frac{pi}{4sqrt{ab}}sum_{m=1}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1} $$
$endgroup$
– Hazem Orabi
Jan 24 at 13:11
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Answer: The series diverges when $sqrt{frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:
$tan{frac{pi x}{2}}$ has period 2 and diverges like $frac{1}{n - x}$ near any odd integer $n$. More precisely, there exists a constant $c_1$ so that for sufficiently small $x$ $$left|tan{frac{pi (n - x)}{2}}right| > left|frac{c_1}{x}right|$$ for any odd $n$.- By the Dirichlet approximation theorem, for irrational $sqrt{frac{b}{a}}$ there are infinitely many integers $m$ such that $sqrt{frac{b}{a}} (2m-1)$ is within distance $frac{c_2}{2m-1}$ of an odd integer, for some positive constant $c_2$.
- Combining the above two remarks, we find that there are infinitely many $m$ such that $$left| frac{tan{frac{pi}{2}sqrt{frac{b}{a}} (2m-1)}}{2m-1} right|>frac{c_1}{c_2}$$
Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.
Remark: If $sqrt{frac{b}{a}}$ is some rational number $frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:
- If $p$ and $q$ are odd, then for some $m$ we find $sqrt{frac{b}{a}} (2m-1)$ will be an odd integer and the tangent in the sum will diverge. Thus the sum doesn't converge.
- If $p$ is odd and $q$ is even, then $sqrt{frac{b}{a}} (2m-1)$ is never an integer, and is symmetrically distributed across the period of the tangent. Note that the tangent is an odd function when reflected across any even integer, so for each negative value of the tangent in the sum there is a corresponding equal and opposite positive value. Within each period of the tangent there are a finite number of such pairs, and the contribution of each such pair to the sum goes like $~frac{1}{m^2}$, which is convergent. (This is exactly analogous to how $sum frac{(-1)^n}{n}$ converges conditionally.) Thus the sum will (conditionally) converge.
- If $p$ is even and $q$ is odd, most points will pair up like in the previous case, but there will be a leftover collection of points $sqrt{frac{b}{a}} (2m-1)$ which land on even integers. Fortunately, the tangent is zero on these points, so the sum will again converge like before.
So summarily the sum converges only when $sqrt{frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $sum frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.
(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)
answered Jan 26 at 21:36
YlyYly
6,78921539
$endgroup$
add a comment |
$begingroup$
Let
$$q=sqrt{dfrac{b}{a}},$$
then attack via digamma-function leads to
$$begin{align}
&S = dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{(2n-1)^2-q^2(2m-1)^2} \[4pt]
&= dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{2(2n-1)}left(dfrac1{2n-1-q(2m-1)}+dfrac1{2n-1+q(2m-1)}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}sumlimits_{m=1}^infty left(-dfrac1{m-frac{2n-1+q}{2q}}+dfrac1{m+frac{2n-1-q}{2q}}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(1-frac{2n-1+q}{2q}right)-psileft(1+frac{2n-1-q}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(frac{q-2n+1}{2q}right)-psileft(frac{q+2n-1}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}cdotpicotleft(pifrac{q-2n+1}{2q}right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{n=1}^inftydfrac1{2n-1}tanleft(fracpi 2 sqrt{frac ab}(2n-1)right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{m=1}^inftydfrac1{2m-1}tanleft(fracpi 2 sqrt{color{red}{frac ab}}(2m-1)right).
end{align}$$
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
$endgroup$
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074662%2fis-there-a-way-to-evaluate-analytically-the-following-infinite-double-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Answer: The series diverges when $sqrt{frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:
$tan{frac{pi x}{2}}$ has period 2 and diverges like $frac{1}{n - x}$ near any odd integer $n$. More precisely, there exists a constant $c_1$ so that for sufficiently small $x$ $$left|tan{frac{pi (n - x)}{2}}right| > left|frac{c_1}{x}right|$$ for any odd $n$.- By the Dirichlet approximation theorem, for irrational $sqrt{frac{b}{a}}$ there are infinitely many integers $m$ such that $sqrt{frac{b}{a}} (2m-1)$ is within distance $frac{c_2}{2m-1}$ of an odd integer, for some positive constant $c_2$.
- Combining the above two remarks, we find that there are infinitely many $m$ such that $$left| frac{tan{frac{pi}{2}sqrt{frac{b}{a}} (2m-1)}}{2m-1} right|>frac{c_1}{c_2}$$
Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.
Remark: If $sqrt{frac{b}{a}}$ is some rational number $frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:
- If $p$ and $q$ are odd, then for some $m$ we find $sqrt{frac{b}{a}} (2m-1)$ will be an odd integer and the tangent in the sum will diverge. Thus the sum doesn't converge.
- If $p$ is odd and $q$ is even, then $sqrt{frac{b}{a}} (2m-1)$ is never an integer, and is symmetrically distributed across the period of the tangent. Note that the tangent is an odd function when reflected across any even integer, so for each negative value of the tangent in the sum there is a corresponding equal and opposite positive value. Within each period of the tangent there are a finite number of such pairs, and the contribution of each such pair to the sum goes like $~frac{1}{m^2}$, which is convergent. (This is exactly analogous to how $sum frac{(-1)^n}{n}$ converges conditionally.) Thus the sum will (conditionally) converge.
- If $p$ is even and $q$ is odd, most points will pair up like in the previous case, but there will be a leftover collection of points $sqrt{frac{b}{a}} (2m-1)$ which land on even integers. Fortunately, the tangent is zero on these points, so the sum will again converge like before.
So summarily the sum converges only when $sqrt{frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $sum frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.
(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)
answered Jan 26 at 21:36
YlyYly
6,78921539
$endgroup$
add a comment |
$begingroup$
Answer: The series diverges when $sqrt{frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:
$tan{frac{pi x}{2}}$ has period 2 and diverges like $frac{1}{n - x}$ near any odd integer $n$. More precisely, there exists a constant $c_1$ so that for sufficiently small $x$ $$left|tan{frac{pi (n - x)}{2}}right| > left|frac{c_1}{x}right|$$ for any odd $n$.- By the Dirichlet approximation theorem, for irrational $sqrt{frac{b}{a}}$ there are infinitely many integers $m$ such that $sqrt{frac{b}{a}} (2m-1)$ is within distance $frac{c_2}{2m-1}$ of an odd integer, for some positive constant $c_2$.
- Combining the above two remarks, we find that there are infinitely many $m$ such that $$left| frac{tan{frac{pi}{2}sqrt{frac{b}{a}} (2m-1)}}{2m-1} right|>frac{c_1}{c_2}$$
Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.
Remark: If $sqrt{frac{b}{a}}$ is some rational number $frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:
- If $p$ and $q$ are odd, then for some $m$ we find $sqrt{frac{b}{a}} (2m-1)$ will be an odd integer and the tangent in the sum will diverge. Thus the sum doesn't converge.
- If $p$ is odd and $q$ is even, then $sqrt{frac{b}{a}} (2m-1)$ is never an integer, and is symmetrically distributed across the period of the tangent. Note that the tangent is an odd function when reflected across any even integer, so for each negative value of the tangent in the sum there is a corresponding equal and opposite positive value. Within each period of the tangent there are a finite number of such pairs, and the contribution of each such pair to the sum goes like $~frac{1}{m^2}$, which is convergent. (This is exactly analogous to how $sum frac{(-1)^n}{n}$ converges conditionally.) Thus the sum will (conditionally) converge.
- If $p$ is even and $q$ is odd, most points will pair up like in the previous case, but there will be a leftover collection of points $sqrt{frac{b}{a}} (2m-1)$ which land on even integers. Fortunately, the tangent is zero on these points, so the sum will again converge like before.
So summarily the sum converges only when $sqrt{frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $sum frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.
(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)
answered Jan 26 at 21:36
YlyYly
6,78921539
$endgroup$
add a comment |
$begingroup$
Answer: The series diverges when $sqrt{frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:
$tan{frac{pi x}{2}}$ has period 2 and diverges like $frac{1}{n - x}$ near any odd integer $n$. More precisely, there exists a constant $c_1$ so that for sufficiently small $x$ $$left|tan{frac{pi (n - x)}{2}}right| > left|frac{c_1}{x}right|$$ for any odd $n$.- By the Dirichlet approximation theorem, for irrational $sqrt{frac{b}{a}}$ there are infinitely many integers $m$ such that $sqrt{frac{b}{a}} (2m-1)$ is within distance $frac{c_2}{2m-1}$ of an odd integer, for some positive constant $c_2$.
- Combining the above two remarks, we find that there are infinitely many $m$ such that $$left| frac{tan{frac{pi}{2}sqrt{frac{b}{a}} (2m-1)}}{2m-1} right|>frac{c_1}{c_2}$$
Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.
Remark: If $sqrt{frac{b}{a}}$ is some rational number $frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:
- If $p$ and $q$ are odd, then for some $m$ we find $sqrt{frac{b}{a}} (2m-1)$ will be an odd integer and the tangent in the sum will diverge. Thus the sum doesn't converge.
- If $p$ is odd and $q$ is even, then $sqrt{frac{b}{a}} (2m-1)$ is never an integer, and is symmetrically distributed across the period of the tangent. Note that the tangent is an odd function when reflected across any even integer, so for each negative value of the tangent in the sum there is a corresponding equal and opposite positive value. Within each period of the tangent there are a finite number of such pairs, and the contribution of each such pair to the sum goes like $~frac{1}{m^2}$, which is convergent. (This is exactly analogous to how $sum frac{(-1)^n}{n}$ converges conditionally.) Thus the sum will (conditionally) converge.
- If $p$ is even and $q$ is odd, most points will pair up like in the previous case, but there will be a leftover collection of points $sqrt{frac{b}{a}} (2m-1)$ which land on even integers. Fortunately, the tangent is zero on these points, so the sum will again converge like before.
So summarily the sum converges only when $sqrt{frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $sum frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.
(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)
answered Jan 26 at 21:36
YlyYly
6,78921539
$endgroup$
Answer: The series diverges when $sqrt{frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:
$tan{frac{pi x}{2}}$ has period 2 and diverges like $frac{1}{n - x}$ near any odd integer $n$. More precisely, there exists a constant $c_1$ so that for sufficiently small $x$ $$left|tan{frac{pi (n - x)}{2}}right| > left|frac{c_1}{x}right|$$ for any odd $n$.- By the Dirichlet approximation theorem, for irrational $sqrt{frac{b}{a}}$ there are infinitely many integers $m$ such that $sqrt{frac{b}{a}} (2m-1)$ is within distance $frac{c_2}{2m-1}$ of an odd integer, for some positive constant $c_2$.
- Combining the above two remarks, we find that there are infinitely many $m$ such that $$left| frac{tan{frac{pi}{2}sqrt{frac{b}{a}} (2m-1)}}{2m-1} right|>frac{c_1}{c_2}$$
Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.
Remark: If $sqrt{frac{b}{a}}$ is some rational number $frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:
- If $p$ and $q$ are odd, then for some $m$ we find $sqrt{frac{b}{a}} (2m-1)$ will be an odd integer and the tangent in the sum will diverge. Thus the sum doesn't converge.
- If $p$ is odd and $q$ is even, then $sqrt{frac{b}{a}} (2m-1)$ is never an integer, and is symmetrically distributed across the period of the tangent. Note that the tangent is an odd function when reflected across any even integer, so for each negative value of the tangent in the sum there is a corresponding equal and opposite positive value. Within each period of the tangent there are a finite number of such pairs, and the contribution of each such pair to the sum goes like $~frac{1}{m^2}$, which is convergent. (This is exactly analogous to how $sum frac{(-1)^n}{n}$ converges conditionally.) Thus the sum will (conditionally) converge.
- If $p$ is even and $q$ is odd, most points will pair up like in the previous case, but there will be a leftover collection of points $sqrt{frac{b}{a}} (2m-1)$ which land on even integers. Fortunately, the tangent is zero on these points, so the sum will again converge like before.
So summarily the sum converges only when $sqrt{frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $sum frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.
(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)
answered Jan 26 at 21:36
YlyYly
6,78921539
edited Jan 26 at 22:09
answered Jan 26 at 21:36
YlyYly
6,78921539
answered Jan 26 at 21:36
YlyYly
6,78921539
answered Jan 26 at 21:36
YlyYly
6,78921539
6,78921539
add a comment |
add a comment |
$begingroup$
Let
$$q=sqrt{dfrac{b}{a}},$$
then attack via digamma-function leads to
$$begin{align}
&S = dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{(2n-1)^2-q^2(2m-1)^2} \[4pt]
&= dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{2(2n-1)}left(dfrac1{2n-1-q(2m-1)}+dfrac1{2n-1+q(2m-1)}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}sumlimits_{m=1}^infty left(-dfrac1{m-frac{2n-1+q}{2q}}+dfrac1{m+frac{2n-1-q}{2q}}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(1-frac{2n-1+q}{2q}right)-psileft(1+frac{2n-1-q}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(frac{q-2n+1}{2q}right)-psileft(frac{q+2n-1}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}cdotpicotleft(pifrac{q-2n+1}{2q}right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{n=1}^inftydfrac1{2n-1}tanleft(fracpi 2 sqrt{frac ab}(2n-1)right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{m=1}^inftydfrac1{2m-1}tanleft(fracpi 2 sqrt{color{red}{frac ab}}(2m-1)right).
end{align}$$
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
$endgroup$
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
add a comment |
$begingroup$
Let
$$q=sqrt{dfrac{b}{a}},$$
then attack via digamma-function leads to
$$begin{align}
&S = dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{(2n-1)^2-q^2(2m-1)^2} \[4pt]
&= dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{2(2n-1)}left(dfrac1{2n-1-q(2m-1)}+dfrac1{2n-1+q(2m-1)}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}sumlimits_{m=1}^infty left(-dfrac1{m-frac{2n-1+q}{2q}}+dfrac1{m+frac{2n-1-q}{2q}}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(1-frac{2n-1+q}{2q}right)-psileft(1+frac{2n-1-q}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(frac{q-2n+1}{2q}right)-psileft(frac{q+2n-1}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}cdotpicotleft(pifrac{q-2n+1}{2q}right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{n=1}^inftydfrac1{2n-1}tanleft(fracpi 2 sqrt{frac ab}(2n-1)right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{m=1}^inftydfrac1{2m-1}tanleft(fracpi 2 sqrt{color{red}{frac ab}}(2m-1)right).
end{align}$$
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
$endgroup$
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
add a comment |
$begingroup$
Let
$$q=sqrt{dfrac{b}{a}},$$
then attack via digamma-function leads to
$$begin{align}
&S = dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{(2n-1)^2-q^2(2m-1)^2} \[4pt]
&= dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{2(2n-1)}left(dfrac1{2n-1-q(2m-1)}+dfrac1{2n-1+q(2m-1)}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}sumlimits_{m=1}^infty left(-dfrac1{m-frac{2n-1+q}{2q}}+dfrac1{m+frac{2n-1-q}{2q}}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(1-frac{2n-1+q}{2q}right)-psileft(1+frac{2n-1-q}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(frac{q-2n+1}{2q}right)-psileft(frac{q+2n-1}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}cdotpicotleft(pifrac{q-2n+1}{2q}right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{n=1}^inftydfrac1{2n-1}tanleft(fracpi 2 sqrt{frac ab}(2n-1)right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{m=1}^inftydfrac1{2m-1}tanleft(fracpi 2 sqrt{color{red}{frac ab}}(2m-1)right).
end{align}$$
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
$endgroup$
Let
$$q=sqrt{dfrac{b}{a}},$$
then attack via digamma-function leads to
$$begin{align}
&S = dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{(2n-1)^2-q^2(2m-1)^2} \[4pt]
&= dfrac1asumlimits_{n=1}^inftysumlimits_{m=1}^inftydfrac1{2(2n-1)}left(dfrac1{2n-1-q(2m-1)}+dfrac1{2n-1+q(2m-1)}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}sumlimits_{m=1}^infty left(-dfrac1{m-frac{2n-1+q}{2q}}+dfrac1{m+frac{2n-1-q}{2q}}right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(1-frac{2n-1+q}{2q}right)-psileft(1+frac{2n-1-q}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}left(psileft(frac{q-2n+1}{2q}right)-psileft(frac{q+2n-1}{2q}right)right)\[4pt]
&= dfrac1{4aq}sumlimits_{n=1}^inftydfrac1{2n-1}cdotpicotleft(pifrac{q-2n+1}{2q}right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{n=1}^inftydfrac1{2n-1}tanleft(fracpi 2 sqrt{frac ab}(2n-1)right)\[4pt]
&= dfracpi{4sqrt{ab}}sumlimits_{m=1}^inftydfrac1{2m-1}tanleft(fracpi 2 sqrt{color{red}{frac ab}}(2m-1)right).
end{align}$$
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
edited Jan 29 at 17:11
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
answered Jan 27 at 23:09
Yuri NegometyanovYuri Negometyanov
11.8k1729
11.8k1729
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
add a comment |
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
(+1) $$ begin{align}S&=color{red}{+}frac{pi}{4sqrt{ab}}sum_{color{red}{m=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1}\&=color{red}{-}frac{pi}{4sqrt{ab}}sum_{color{red}{n=1}}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{a}{b}}}left(2n-1right)right)}{2n-1}end{align} $$
$endgroup$
– Hazem Orabi
Jan 29 at 13:33
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
$begingroup$
@HazemOrabi I've elaborated my reasons in the answer. What are yours?
$endgroup$
– Yuri Negometyanov
Jan 29 at 17:14
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074662%2fis-there-a-way-to-evaluate-analytically-the-following-infinite-double-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Maybe Poisson summation formula?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:28
$begingroup$
@LordSharktheUnknown Thanks for the comments. Could you please be a bit specific. An example would be highly appreciated
$endgroup$
– Math Student
Jan 15 at 17:31
4
$begingroup$
On second thoughts, I believe now that the double sum is divergent.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 18:07
1
$begingroup$
@LordSharktheUnknown i agree. The divergence looks logarithmic. Thanks for the hint
$endgroup$
– Math Student
Jan 16 at 7:25
2
$begingroup$
$$ S=frac{pi}{4sqrt{ab}}sum_{m=1}^inftyfrac{tanleft(frac{pi}{2}sqrt{color{red}{frac{b}{a}}}left(2m-1right)right)}{2m-1} $$
$endgroup$
– Hazem Orabi
Jan 24 at 13:11