Mixing
Infinite series problem
$begingroup$
The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$
Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
sequences-and-series
asked Jan 24 at 13:00
DXTDXT
6,0612732
$endgroup$
add a comment |
$begingroup$
The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$
Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
sequences-and-series
asked Jan 24 at 13:00
DXTDXT
6,0612732
$endgroup$
1
$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07
$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09
add a comment |
$begingroup$
The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$
Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
sequences-and-series
asked Jan 24 at 13:00
DXTDXT
6,0612732
$endgroup$
The sum of $$frac{2}{4-1}+frac{2^2}{4^2-1}+frac{2^4}{4^4-1}+cdots cdots $$
Try: write it as $$S = sum^{infty}_{r=0}frac{2^{2^{r}}}{2^{2^{r+1}}-1}=sum^{infty}_{r=0}frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
sequences-and-series
sequences-and-series
asked Jan 24 at 13:00
DXTDXT
6,0612732
asked Jan 24 at 13:00
DXTDXT
6,0612732
asked Jan 24 at 13:00
DXTDXT
6,0612732
asked Jan 24 at 13:00
DXTDXT
6,0612732
asked Jan 24 at 13:00
DXTDXT
6,0612732
6,0612732
1
$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07
$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09
add a comment |
1
$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07
$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09
1
1
$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07
$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07
$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09
$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You may continue as:
$$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$
After cancelling the terms you are left with:
$$1 - frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
$endgroup$
1
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
1
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is2^{2^{j+1}}, not2^2^{j+1}. That's why your comment malfunctioned.
$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
add a comment |
$begingroup$
You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
$endgroup$
add a comment |
$begingroup$
Note first that
$$
frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
$$ If we sum over $n=2^j$, $jge 0$, we have
$$
S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
$$ since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.
answered Jan 24 at 13:11
SongSong
18k21449
$endgroup$
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%2f3085853%2finfinite-series-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You may continue as:
$$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$
After cancelling the terms you are left with:
$$1 - frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
$endgroup$
1
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
1
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is2^{2^{j+1}}, not2^2^{j+1}. That's why your comment malfunctioned.
$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
add a comment |
$begingroup$
You may continue as:
$$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$
After cancelling the terms you are left with:
$$1 - frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
$endgroup$
1
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
1
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is2^{2^{j+1}}, not2^2^{j+1}. That's why your comment malfunctioned.
$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
add a comment |
$begingroup$
You may continue as:
$$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$
After cancelling the terms you are left with:
$$1 - frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
$endgroup$
You may continue as:
$$frac{2^n}{(2^n-1)(2^n+1)} =frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$left(frac 11 - frac 13right) + left(frac 13 - frac 1{15}right) + ... + left(frac 1{2^n-1} - frac 1{(2^n-1)(2^n+1)}right)$$
After cancelling the terms you are left with:
$$1 - frac 1{(2^n-1)(2^n+1)}$$
When $n$ tends to infinity, the expression becomes $1$.
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
edited Jan 25 at 0:21
Paul Sinclair
20.5k21543
20.5k21543
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
answered Jan 24 at 16:02
AspiringEngineerAspiringEngineer
514
514
1
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
1
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is2^{2^{j+1}}, not2^2^{j+1}. That's why your comment malfunctioned.
$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
add a comment |
1
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
1
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is2^{2^{j+1}}, not2^2^{j+1}. That's why your comment malfunctioned.
$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
1
1
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
$begingroup$
Welcome to Math.SE. Note that mathematical notation can be properly typeset in posts, which makes them easier to read.
$endgroup$
– hardmath
Jan 24 at 16:40
1
1
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@PaulSinclair In the solution , $n=2^j$ , so the pattern does indeed follow , as $$frac{1}{2^2^{j+1}-1}=frac{1}{(2^2^j+1)(2^2^j-1)}$$ . While the solution should have mentioned this , I hope this clears it up .
$endgroup$
– Sinπ
Jan 24 at 19:05
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is
2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@Rahuboy For future reference you can't do double exponentials without braces, e.g. $2^{2^{j+1}}$ is
2^{2^{j+1}}, not 2^2^{j+1}. That's why your comment malfunctioned.$endgroup$
– J.G.
Jan 24 at 22:29
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
@J.G. Oh I see.... thanks for the help !
$endgroup$
– Sinπ
Jan 25 at 0:02
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
$begingroup$
That is embarrassing. My apologies for failing to realize this.
$endgroup$
– Paul Sinclair
Jan 25 at 0:14
add a comment |
$begingroup$
You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
$endgroup$
add a comment |
$begingroup$
You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
$endgroup$
add a comment |
$begingroup$
You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
$endgroup$
You may prove by induction that $$sum_{r=0}^nfrac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{n+1}}-1}.$$(If yoiu didn't spot this conjecture at first, you will after calculating the first few partial sums.) Indeed the claim is correct if $n=0$, and if it holds for $n=k$ then $$sum_{r=0}^{k+1}frac{2^{2^r}}{2^{2^{r+1}}-1}=1-frac{1}{2^{2^{k+1}}-1}+frac{2^{2^{k+1}}}{2^{2^{k+2}}-1}=1-frac{1}{2^{2^{k+2}}-1}$$as required, where the final calculation is the $a=2^{2^{k+1}}$ special case of $$1-frac{1}{a-1}+frac{a}{a^2-1}=1-frac{1}{a^2-1}.$$You may wish to rewrite this argument as the computation of a telescoping series.
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
answered Jan 24 at 13:13
J.G.J.G.
30.3k23148
30.3k23148
add a comment |
add a comment |
$begingroup$
Note first that
$$
frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
$$ If we sum over $n=2^j$, $jge 0$, we have
$$
S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
$$ since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.
answered Jan 24 at 13:11
SongSong
18k21449
$endgroup$
add a comment |
$begingroup$
Note first that
$$
frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
$$ If we sum over $n=2^j$, $jge 0$, we have
$$
S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
$$ since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.
answered Jan 24 at 13:11
SongSong
18k21449
$endgroup$
add a comment |
$begingroup$
Note first that
$$
frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
$$ If we sum over $n=2^j$, $jge 0$, we have
$$
S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
$$ since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.
answered Jan 24 at 13:11
SongSong
18k21449
$endgroup$
Note first that
$$
frac{2^n}{2^{2n}-1}=2^{-n}frac{1}{1-2^{-2n}}=sum_{k:text{odd},kinBbb N} 2^{-nk}.
$$ If we sum over $n=2^j$, $jge 0$, we have
$$
S=sum_{j=0}^infty frac{2^{2^j}}{2^{2^{j+1}}-1}=sum_{j=0}^inftysum_{k:text{odd},kinBbb N} 2^{-2^jcdot k}=sum_{l=1}^infty 2^{-l}=1
$$ since every $lge 1$ has a unique representation $l=2^jcdot k$ for some $jge 0$ and odd $kin Bbb N$.
answered Jan 24 at 13:11
SongSong
18k21449
edited Jan 24 at 13:19
answered Jan 24 at 13:11
SongSong
18k21449
answered Jan 24 at 13:11
SongSong
18k21449
answered Jan 24 at 13:11
SongSong
18k21449
18k21449
add a comment |
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%2f3085853%2finfinite-series-problem%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
1
$begingroup$
$$dfrac a{a^2-1}-dfrac a{a^2+1}=dfrac 2{a^4-1}$$
$endgroup$
– lab bhattacharjee
Jan 24 at 13:07
$begingroup$
@labbhattacharjee I think he had the right side and wrote the left one...
$endgroup$
– DonAntonio
Jan 24 at 13:09