Mixing
How to take every third element in a series?
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$sum_{n=1}^infty frac{1}{n^2} = frac{1}{1^2} + frac{1}{2^2} + frac{1}{3^2} + cdots = 1.644934$ or $frac{pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ frac{1}{3^2} + frac{1}{6^2} + frac{1}{9^2} + cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ frac{1}{2^2} + frac{1}{5^2} + frac{1}{8^2} + cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ frac{1}{1^2} + frac{1}{4^2} + frac{1}{7^2} + cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
power-series
asked Jan 3 at 16:25
DaleDale
1,1801334
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add a comment |
$begingroup$
$sum_{n=1}^infty frac{1}{n^2} = frac{1}{1^2} + frac{1}{2^2} + frac{1}{3^2} + cdots = 1.644934$ or $frac{pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ frac{1}{3^2} + frac{1}{6^2} + frac{1}{9^2} + cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ frac{1}{2^2} + frac{1}{5^2} + frac{1}{8^2} + cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ frac{1}{1^2} + frac{1}{4^2} + frac{1}{7^2} + cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
power-series
asked Jan 3 at 16:25
DaleDale
1,1801334
$endgroup$
1
$begingroup$
The first sum you mentioned is $frac{1}{9}$ of the initial sum.
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– Peter
Jan 3 at 16:26
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The first sum can be rewritten as$$frac 19+frac 1{36}+cdots=frac 19left[1+frac 14+frac 19+cdotsright]color{blue}{=frac 19zeta(2)}$$
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– Frank W.
Jan 3 at 16:30
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If I may, but what is this problem for? When would you ever need to calculate each sum?
$endgroup$
– Frank W.
Jan 3 at 16:31
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@Peter that means that the other two sums must be different. as if they were the same the total would be $frac{3}{9}$ the initial sum.
$endgroup$
– Dale
Jan 3 at 16:38
$begingroup$
@Dale Peter said the first sum
$endgroup$
– Frank W.
Jan 3 at 16:43
add a comment |
$begingroup$
$sum_{n=1}^infty frac{1}{n^2} = frac{1}{1^2} + frac{1}{2^2} + frac{1}{3^2} + cdots = 1.644934$ or $frac{pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ frac{1}{3^2} + frac{1}{6^2} + frac{1}{9^2} + cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ frac{1}{2^2} + frac{1}{5^2} + frac{1}{8^2} + cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ frac{1}{1^2} + frac{1}{4^2} + frac{1}{7^2} + cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
power-series
asked Jan 3 at 16:25
DaleDale
1,1801334
$endgroup$
$sum_{n=1}^infty frac{1}{n^2} = frac{1}{1^2} + frac{1}{2^2} + frac{1}{3^2} + cdots = 1.644934$ or $frac{pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ frac{1}{3^2} + frac{1}{6^2} + frac{1}{9^2} + cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ frac{1}{2^2} + frac{1}{5^2} + frac{1}{8^2} + cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ frac{1}{1^2} + frac{1}{4^2} + frac{1}{7^2} + cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
power-series
power-series
asked Jan 3 at 16:25
DaleDale
1,1801334
asked Jan 3 at 16:25
DaleDale
1,1801334
edited Jan 13 at 19:36
Dale
asked Jan 3 at 16:25
DaleDale
1,1801334
asked Jan 3 at 16:25
DaleDale
1,1801334
asked Jan 3 at 16:25
DaleDale
1,1801334
1,1801334
1
$begingroup$
The first sum you mentioned is $frac{1}{9}$ of the initial sum.
$endgroup$
– Peter
Jan 3 at 16:26
$begingroup$
The first sum can be rewritten as$$frac 19+frac 1{36}+cdots=frac 19left[1+frac 14+frac 19+cdotsright]color{blue}{=frac 19zeta(2)}$$
$endgroup$
– Frank W.
Jan 3 at 16:30
$begingroup$
If I may, but what is this problem for? When would you ever need to calculate each sum?
$endgroup$
– Frank W.
Jan 3 at 16:31
$begingroup$
@Peter that means that the other two sums must be different. as if they were the same the total would be $frac{3}{9}$ the initial sum.
$endgroup$
– Dale
Jan 3 at 16:38
$begingroup$
@Dale Peter said the first sum
$endgroup$
– Frank W.
Jan 3 at 16:43
add a comment |
1
$begingroup$
The first sum you mentioned is $frac{1}{9}$ of the initial sum.
$endgroup$
– Peter
Jan 3 at 16:26
$begingroup$
The first sum can be rewritten as$$frac 19+frac 1{36}+cdots=frac 19left[1+frac 14+frac 19+cdotsright]color{blue}{=frac 19zeta(2)}$$
$endgroup$
– Frank W.
Jan 3 at 16:30
$begingroup$
If I may, but what is this problem for? When would you ever need to calculate each sum?
$endgroup$
– Frank W.
Jan 3 at 16:31
$begingroup$
@Peter that means that the other two sums must be different. as if they were the same the total would be $frac{3}{9}$ the initial sum.
$endgroup$
– Dale
Jan 3 at 16:38
$begingroup$
@Dale Peter said the first sum
$endgroup$
– Frank W.
Jan 3 at 16:43
1
1
$begingroup$
The first sum you mentioned is $frac{1}{9}$ of the initial sum.
$endgroup$
– Peter
Jan 3 at 16:26
$begingroup$
The first sum you mentioned is $frac{1}{9}$ of the initial sum.
$endgroup$
– Peter
Jan 3 at 16:26
$begingroup$
The first sum can be rewritten as$$frac 19+frac 1{36}+cdots=frac 19left[1+frac 14+frac 19+cdotsright]color{blue}{=frac 19zeta(2)}$$
$endgroup$
– Frank W.
Jan 3 at 16:30
$begingroup$
The first sum can be rewritten as$$frac 19+frac 1{36}+cdots=frac 19left[1+frac 14+frac 19+cdotsright]color{blue}{=frac 19zeta(2)}$$
$endgroup$
– Frank W.
Jan 3 at 16:30
$begingroup$
If I may, but what is this problem for? When would you ever need to calculate each sum?
$endgroup$
– Frank W.
Jan 3 at 16:31
$begingroup$
If I may, but what is this problem for? When would you ever need to calculate each sum?
$endgroup$
– Frank W.
Jan 3 at 16:31
$begingroup$
@Peter that means that the other two sums must be different. as if they were the same the total would be $frac{3}{9}$ the initial sum.
$endgroup$
– Dale
Jan 3 at 16:38
$begingroup$
@Peter that means that the other two sums must be different. as if they were the same the total would be $frac{3}{9}$ the initial sum.
$endgroup$
– Dale
Jan 3 at 16:38
$begingroup$
@Dale Peter said the first sum
$endgroup$
– Frank W.
Jan 3 at 16:43
$begingroup$
@Dale Peter said the first sum
$endgroup$
– Frank W.
Jan 3 at 16:43
add a comment |
4 Answers
4
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oldest
votes
$begingroup$
Note that we have
$$psi'(z)=sum_{n=0}^infty frac{1}{(n+z)^2}$$
where $psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$sum_{n=0}^infty frac{1}{(3n+1)^2}=frac19 psi'(1/3)$$
and
$$sum_{n=0}^infty frac{1}{(3n+2)^2}=frac19 psi'(2/3)$$
Interestingly, since we have
$$sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$$
we find that
$$psi'(1/3)+psi'(2/3) = 4pi^2/3$$
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
$endgroup$
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Why does one psi lack a '
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– Dale
Jan 13 at 18:13
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@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
add a comment |
$begingroup$
As a complement to Mark's answer,
$$sum_{ngeq 0}frac{1}{(3n+1)^2}=-int_{0}^{1}sum_{ngeq 0} x^{3n}log(x),dx=int_{0}^{1}frac{-log x}{1-x^3},dx $$
(and similarly $sum_{ngeq 0}frac{1}{(3n+2)^2}$) can be expressed in terms of dilogarithms, since
$$ int_{0}^{1}frac{-log x}{1-a x}=frac{text{Li}_2(a)}{a} $$
for any $|a|leq 1$, with $text{Li}_2(a)=sum_{ngeq 1}frac{a^n}{n^2}$. This is equivalent to stating that $psi'left(frac{1}{3}right)$ and $psi'left(frac{2}{3}right)$ can be computed through the discrete Fourier transform. It is worth noticing that
$$text{Re},text{Li}_2(e^{itheta})=sum_{ngeq 1}frac{cos(ntheta)}{n^2} $$
is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $text{Im},text{Li}_2(e^{itheta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
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add a comment |
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Polylogarithms
A useful formula that can be applied here is
$$
frac13sum_{k=0}^2e^{2pi ijk/3}=[3mid j]tag1
$$
So
$$
begin{align}newcommand{Li}{operatorname{Li}}
sum_{j=0}^inftyfrac1{(3j+1)^2}
&=frac13sum_{k=0}^2sum_{j=1}^infty e^{2pi i(j-1)k/3}frac1{j^2}\
&=frac13left(frac{pi^2}6+e^{-2pi i/3}Li_2left(e^{2pi i/3}right)+e^{2pi i/3}Li_2left(e^{-2pi i/3}right)right)tag2
end{align}
$$
Mathematica gives $1.12173301393634378687$ usingN[1/3(Pi^2/6 + Exp[-2Pi I/3]PolyLog[2,Exp[2Pi I/3]]+
Exp[2Pi I/3] PolyLog[2,Exp[-2Pi I/3]]),20]
Extended Harmonic Numbers
Another approach is to use the Extended Harmonic Numbers.
$$
H(x)=sum_{k=1}^inftyleft(frac1k-frac1{k+x}right)tag3
$$
where
$$
H'(x)=sum_{k=1}^inftyfrac1{(k+x)^2}tag5
$$
Giving
$$
frac19H'!left(-frac23right)=sum_{k=0}^inftyfrac1{(3k+1)^2}tag6
$$
Mathematica gives $1.1217330139363437869$ usingN[1/9HarmonicNumber'[-2/3],20]
Euler-Maclaurin Sum Formula
Although it does not give a closed form, the Euler-Maclaurin Sum Formula allows us to accelerate the computation of the sum.
$$
begin{align}
sum_{k=0}^nfrac1{(3k+1)^2}
&sim C-frac1{3(3n+1)}+frac1{2(3n+1)^2}-frac1{2(3n+1)^3}+frac9{10(3n+1)^5}\
&-frac{81}{14(3n+1)^7}+frac{729}{10(3n+1)^9}-frac{32805}{22(3n+1)^{11}}tag7
end{align}
$$
Using $n=100$ in $(7)$, we get
$$
sum_{k=0}^inftyfrac1{(3k+1)^2}=1.1217330139363437868657782tag8
$$
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
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add a comment |
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KM101 deleted his hint... not sure why.
$frac{1}{3^2}+frac{1}{6^2}+frac{1}{9^2}+dots=frac{1}{3^2 1^2}+frac{1}{3^2 2^2}+frac{1}{3^2 3^3}+dots=frac{1}{9}(frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+dots)= frac{pi^2}{54}$
$frac{1}{2^2}+frac{1}{5^2}+frac{1}{8^2}+dots=frac{1}{2^2 1^2}+frac{1}{3^2?? 1^2}+frac{1}{3^2 ??1^2}$
answered Jan 3 at 16:37
DaleDale
1,1801334
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1
$begingroup$
How does this answer the OP's question?
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– Mark Viola
Jan 3 at 20:24
add a comment |
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4 Answers
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4 Answers
4
active
oldest
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$begingroup$
Note that we have
$$psi'(z)=sum_{n=0}^infty frac{1}{(n+z)^2}$$
where $psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$sum_{n=0}^infty frac{1}{(3n+1)^2}=frac19 psi'(1/3)$$
and
$$sum_{n=0}^infty frac{1}{(3n+2)^2}=frac19 psi'(2/3)$$
Interestingly, since we have
$$sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$$
we find that
$$psi'(1/3)+psi'(2/3) = 4pi^2/3$$
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Why does one psi lack a '
$endgroup$
– Dale
Jan 13 at 18:13
$begingroup$
@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
add a comment |
$begingroup$
Note that we have
$$psi'(z)=sum_{n=0}^infty frac{1}{(n+z)^2}$$
where $psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$sum_{n=0}^infty frac{1}{(3n+1)^2}=frac19 psi'(1/3)$$
and
$$sum_{n=0}^infty frac{1}{(3n+2)^2}=frac19 psi'(2/3)$$
Interestingly, since we have
$$sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$$
we find that
$$psi'(1/3)+psi'(2/3) = 4pi^2/3$$
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Why does one psi lack a '
$endgroup$
– Dale
Jan 13 at 18:13
$begingroup$
@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
add a comment |
$begingroup$
Note that we have
$$psi'(z)=sum_{n=0}^infty frac{1}{(n+z)^2}$$
where $psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$sum_{n=0}^infty frac{1}{(3n+1)^2}=frac19 psi'(1/3)$$
and
$$sum_{n=0}^infty frac{1}{(3n+2)^2}=frac19 psi'(2/3)$$
Interestingly, since we have
$$sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$$
we find that
$$psi'(1/3)+psi'(2/3) = 4pi^2/3$$
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
$endgroup$
Note that we have
$$psi'(z)=sum_{n=0}^infty frac{1}{(n+z)^2}$$
where $psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$sum_{n=0}^infty frac{1}{(3n+1)^2}=frac19 psi'(1/3)$$
and
$$sum_{n=0}^infty frac{1}{(3n+2)^2}=frac19 psi'(2/3)$$
Interestingly, since we have
$$sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$$
we find that
$$psi'(1/3)+psi'(2/3) = 4pi^2/3$$
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
edited Jan 13 at 18:59
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
answered Jan 3 at 17:02
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
Why does one psi lack a '
$endgroup$
– Dale
Jan 13 at 18:13
$begingroup$
@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
add a comment |
$begingroup$
Why does one psi lack a '
$endgroup$
– Dale
Jan 13 at 18:13
$begingroup$
@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
$begingroup$
Why does one psi lack a '
$endgroup$
– Dale
Jan 13 at 18:13
$begingroup$
Why does one psi lack a '
$endgroup$
– Dale
Jan 13 at 18:13
$begingroup$
@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
@dale That ommission is an obvious typographical error. I've edited accordingly.
$endgroup$
– Mark Viola
Jan 13 at 18:59
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
$begingroup$
Yes, at least one of us. Note that $sum_{n=0}^infty left(frac1{(3n+3)^2}+frac1{(3n+2)^2}+frac1{(3n+1)^2}right)=frac{pi^2}{6}$ and hence $$frac19(psi'(1/3)+psi'(2/3))=frac{pi^2}{6}-frac{pi^2}{54}=frac{8pi^2}{54}=frac19left(frac{4pi^2}{3}right)$$Now multiply by $9$.
$endgroup$
– Mark Viola
Jan 13 at 19:29
add a comment |
$begingroup$
As a complement to Mark's answer,
$$sum_{ngeq 0}frac{1}{(3n+1)^2}=-int_{0}^{1}sum_{ngeq 0} x^{3n}log(x),dx=int_{0}^{1}frac{-log x}{1-x^3},dx $$
(and similarly $sum_{ngeq 0}frac{1}{(3n+2)^2}$) can be expressed in terms of dilogarithms, since
$$ int_{0}^{1}frac{-log x}{1-a x}=frac{text{Li}_2(a)}{a} $$
for any $|a|leq 1$, with $text{Li}_2(a)=sum_{ngeq 1}frac{a^n}{n^2}$. This is equivalent to stating that $psi'left(frac{1}{3}right)$ and $psi'left(frac{2}{3}right)$ can be computed through the discrete Fourier transform. It is worth noticing that
$$text{Re},text{Li}_2(e^{itheta})=sum_{ngeq 1}frac{cos(ntheta)}{n^2} $$
is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $text{Im},text{Li}_2(e^{itheta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
As a complement to Mark's answer,
$$sum_{ngeq 0}frac{1}{(3n+1)^2}=-int_{0}^{1}sum_{ngeq 0} x^{3n}log(x),dx=int_{0}^{1}frac{-log x}{1-x^3},dx $$
(and similarly $sum_{ngeq 0}frac{1}{(3n+2)^2}$) can be expressed in terms of dilogarithms, since
$$ int_{0}^{1}frac{-log x}{1-a x}=frac{text{Li}_2(a)}{a} $$
for any $|a|leq 1$, with $text{Li}_2(a)=sum_{ngeq 1}frac{a^n}{n^2}$. This is equivalent to stating that $psi'left(frac{1}{3}right)$ and $psi'left(frac{2}{3}right)$ can be computed through the discrete Fourier transform. It is worth noticing that
$$text{Re},text{Li}_2(e^{itheta})=sum_{ngeq 1}frac{cos(ntheta)}{n^2} $$
is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $text{Im},text{Li}_2(e^{itheta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
As a complement to Mark's answer,
$$sum_{ngeq 0}frac{1}{(3n+1)^2}=-int_{0}^{1}sum_{ngeq 0} x^{3n}log(x),dx=int_{0}^{1}frac{-log x}{1-x^3},dx $$
(and similarly $sum_{ngeq 0}frac{1}{(3n+2)^2}$) can be expressed in terms of dilogarithms, since
$$ int_{0}^{1}frac{-log x}{1-a x}=frac{text{Li}_2(a)}{a} $$
for any $|a|leq 1$, with $text{Li}_2(a)=sum_{ngeq 1}frac{a^n}{n^2}$. This is equivalent to stating that $psi'left(frac{1}{3}right)$ and $psi'left(frac{2}{3}right)$ can be computed through the discrete Fourier transform. It is worth noticing that
$$text{Re},text{Li}_2(e^{itheta})=sum_{ngeq 1}frac{cos(ntheta)}{n^2} $$
is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $text{Im},text{Li}_2(e^{itheta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
As a complement to Mark's answer,
$$sum_{ngeq 0}frac{1}{(3n+1)^2}=-int_{0}^{1}sum_{ngeq 0} x^{3n}log(x),dx=int_{0}^{1}frac{-log x}{1-x^3},dx $$
(and similarly $sum_{ngeq 0}frac{1}{(3n+2)^2}$) can be expressed in terms of dilogarithms, since
$$ int_{0}^{1}frac{-log x}{1-a x}=frac{text{Li}_2(a)}{a} $$
for any $|a|leq 1$, with $text{Li}_2(a)=sum_{ngeq 1}frac{a^n}{n^2}$. This is equivalent to stating that $psi'left(frac{1}{3}right)$ and $psi'left(frac{2}{3}right)$ can be computed through the discrete Fourier transform. It is worth noticing that
$$text{Re},text{Li}_2(e^{itheta})=sum_{ngeq 1}frac{cos(ntheta)}{n^2} $$
is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $text{Im},text{Li}_2(e^{itheta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 20:40
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
$begingroup$
Polylogarithms
A useful formula that can be applied here is
$$
frac13sum_{k=0}^2e^{2pi ijk/3}=[3mid j]tag1
$$
So
$$
begin{align}newcommand{Li}{operatorname{Li}}
sum_{j=0}^inftyfrac1{(3j+1)^2}
&=frac13sum_{k=0}^2sum_{j=1}^infty e^{2pi i(j-1)k/3}frac1{j^2}\
&=frac13left(frac{pi^2}6+e^{-2pi i/3}Li_2left(e^{2pi i/3}right)+e^{2pi i/3}Li_2left(e^{-2pi i/3}right)right)tag2
end{align}
$$
Mathematica gives $1.12173301393634378687$ usingN[1/3(Pi^2/6 + Exp[-2Pi I/3]PolyLog[2,Exp[2Pi I/3]]+
Exp[2Pi I/3] PolyLog[2,Exp[-2Pi I/3]]),20]
Extended Harmonic Numbers
Another approach is to use the Extended Harmonic Numbers.
$$
H(x)=sum_{k=1}^inftyleft(frac1k-frac1{k+x}right)tag3
$$
where
$$
H'(x)=sum_{k=1}^inftyfrac1{(k+x)^2}tag5
$$
Giving
$$
frac19H'!left(-frac23right)=sum_{k=0}^inftyfrac1{(3k+1)^2}tag6
$$
Mathematica gives $1.1217330139363437869$ usingN[1/9HarmonicNumber'[-2/3],20]
Euler-Maclaurin Sum Formula
Although it does not give a closed form, the Euler-Maclaurin Sum Formula allows us to accelerate the computation of the sum.
$$
begin{align}
sum_{k=0}^nfrac1{(3k+1)^2}
&sim C-frac1{3(3n+1)}+frac1{2(3n+1)^2}-frac1{2(3n+1)^3}+frac9{10(3n+1)^5}\
&-frac{81}{14(3n+1)^7}+frac{729}{10(3n+1)^9}-frac{32805}{22(3n+1)^{11}}tag7
end{align}
$$
Using $n=100$ in $(7)$, we get
$$
sum_{k=0}^inftyfrac1{(3k+1)^2}=1.1217330139363437868657782tag8
$$
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
Polylogarithms
A useful formula that can be applied here is
$$
frac13sum_{k=0}^2e^{2pi ijk/3}=[3mid j]tag1
$$
So
$$
begin{align}newcommand{Li}{operatorname{Li}}
sum_{j=0}^inftyfrac1{(3j+1)^2}
&=frac13sum_{k=0}^2sum_{j=1}^infty e^{2pi i(j-1)k/3}frac1{j^2}\
&=frac13left(frac{pi^2}6+e^{-2pi i/3}Li_2left(e^{2pi i/3}right)+e^{2pi i/3}Li_2left(e^{-2pi i/3}right)right)tag2
end{align}
$$
Mathematica gives $1.12173301393634378687$ usingN[1/3(Pi^2/6 + Exp[-2Pi I/3]PolyLog[2,Exp[2Pi I/3]]+
Exp[2Pi I/3] PolyLog[2,Exp[-2Pi I/3]]),20]
Extended Harmonic Numbers
Another approach is to use the Extended Harmonic Numbers.
$$
H(x)=sum_{k=1}^inftyleft(frac1k-frac1{k+x}right)tag3
$$
where
$$
H'(x)=sum_{k=1}^inftyfrac1{(k+x)^2}tag5
$$
Giving
$$
frac19H'!left(-frac23right)=sum_{k=0}^inftyfrac1{(3k+1)^2}tag6
$$
Mathematica gives $1.1217330139363437869$ usingN[1/9HarmonicNumber'[-2/3],20]
Euler-Maclaurin Sum Formula
Although it does not give a closed form, the Euler-Maclaurin Sum Formula allows us to accelerate the computation of the sum.
$$
begin{align}
sum_{k=0}^nfrac1{(3k+1)^2}
&sim C-frac1{3(3n+1)}+frac1{2(3n+1)^2}-frac1{2(3n+1)^3}+frac9{10(3n+1)^5}\
&-frac{81}{14(3n+1)^7}+frac{729}{10(3n+1)^9}-frac{32805}{22(3n+1)^{11}}tag7
end{align}
$$
Using $n=100$ in $(7)$, we get
$$
sum_{k=0}^inftyfrac1{(3k+1)^2}=1.1217330139363437868657782tag8
$$
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
Polylogarithms
A useful formula that can be applied here is
$$
frac13sum_{k=0}^2e^{2pi ijk/3}=[3mid j]tag1
$$
So
$$
begin{align}newcommand{Li}{operatorname{Li}}
sum_{j=0}^inftyfrac1{(3j+1)^2}
&=frac13sum_{k=0}^2sum_{j=1}^infty e^{2pi i(j-1)k/3}frac1{j^2}\
&=frac13left(frac{pi^2}6+e^{-2pi i/3}Li_2left(e^{2pi i/3}right)+e^{2pi i/3}Li_2left(e^{-2pi i/3}right)right)tag2
end{align}
$$
Mathematica gives $1.12173301393634378687$ usingN[1/3(Pi^2/6 + Exp[-2Pi I/3]PolyLog[2,Exp[2Pi I/3]]+
Exp[2Pi I/3] PolyLog[2,Exp[-2Pi I/3]]),20]
Extended Harmonic Numbers
Another approach is to use the Extended Harmonic Numbers.
$$
H(x)=sum_{k=1}^inftyleft(frac1k-frac1{k+x}right)tag3
$$
where
$$
H'(x)=sum_{k=1}^inftyfrac1{(k+x)^2}tag5
$$
Giving
$$
frac19H'!left(-frac23right)=sum_{k=0}^inftyfrac1{(3k+1)^2}tag6
$$
Mathematica gives $1.1217330139363437869$ usingN[1/9HarmonicNumber'[-2/3],20]
Euler-Maclaurin Sum Formula
Although it does not give a closed form, the Euler-Maclaurin Sum Formula allows us to accelerate the computation of the sum.
$$
begin{align}
sum_{k=0}^nfrac1{(3k+1)^2}
&sim C-frac1{3(3n+1)}+frac1{2(3n+1)^2}-frac1{2(3n+1)^3}+frac9{10(3n+1)^5}\
&-frac{81}{14(3n+1)^7}+frac{729}{10(3n+1)^9}-frac{32805}{22(3n+1)^{11}}tag7
end{align}
$$
Using $n=100$ in $(7)$, we get
$$
sum_{k=0}^inftyfrac1{(3k+1)^2}=1.1217330139363437868657782tag8
$$
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
$endgroup$
Polylogarithms
A useful formula that can be applied here is
$$
frac13sum_{k=0}^2e^{2pi ijk/3}=[3mid j]tag1
$$
So
$$
begin{align}newcommand{Li}{operatorname{Li}}
sum_{j=0}^inftyfrac1{(3j+1)^2}
&=frac13sum_{k=0}^2sum_{j=1}^infty e^{2pi i(j-1)k/3}frac1{j^2}\
&=frac13left(frac{pi^2}6+e^{-2pi i/3}Li_2left(e^{2pi i/3}right)+e^{2pi i/3}Li_2left(e^{-2pi i/3}right)right)tag2
end{align}
$$
Mathematica gives $1.12173301393634378687$ usingN[1/3(Pi^2/6 + Exp[-2Pi I/3]PolyLog[2,Exp[2Pi I/3]]+
Exp[2Pi I/3] PolyLog[2,Exp[-2Pi I/3]]),20]
Extended Harmonic Numbers
Another approach is to use the Extended Harmonic Numbers.
$$
H(x)=sum_{k=1}^inftyleft(frac1k-frac1{k+x}right)tag3
$$
where
$$
H'(x)=sum_{k=1}^inftyfrac1{(k+x)^2}tag5
$$
Giving
$$
frac19H'!left(-frac23right)=sum_{k=0}^inftyfrac1{(3k+1)^2}tag6
$$
Mathematica gives $1.1217330139363437869$ usingN[1/9HarmonicNumber'[-2/3],20]
Euler-Maclaurin Sum Formula
Although it does not give a closed form, the Euler-Maclaurin Sum Formula allows us to accelerate the computation of the sum.
$$
begin{align}
sum_{k=0}^nfrac1{(3k+1)^2}
&sim C-frac1{3(3n+1)}+frac1{2(3n+1)^2}-frac1{2(3n+1)^3}+frac9{10(3n+1)^5}\
&-frac{81}{14(3n+1)^7}+frac{729}{10(3n+1)^9}-frac{32805}{22(3n+1)^{11}}tag7
end{align}
$$
Using $n=100$ in $(7)$, we get
$$
sum_{k=0}^inftyfrac1{(3k+1)^2}=1.1217330139363437868657782tag8
$$
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
edited Jan 14 at 3:19
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
answered Jan 13 at 20:48
robjohn♦robjohn
266k27304626
266k27304626
add a comment |
add a comment |
$begingroup$
KM101 deleted his hint... not sure why.
$frac{1}{3^2}+frac{1}{6^2}+frac{1}{9^2}+dots=frac{1}{3^2 1^2}+frac{1}{3^2 2^2}+frac{1}{3^2 3^3}+dots=frac{1}{9}(frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+dots)= frac{pi^2}{54}$
$frac{1}{2^2}+frac{1}{5^2}+frac{1}{8^2}+dots=frac{1}{2^2 1^2}+frac{1}{3^2?? 1^2}+frac{1}{3^2 ??1^2}$
answered Jan 3 at 16:37
DaleDale
1,1801334
$endgroup$
1
$begingroup$
How does this answer the OP's question?
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
$begingroup$
KM101 deleted his hint... not sure why.
$frac{1}{3^2}+frac{1}{6^2}+frac{1}{9^2}+dots=frac{1}{3^2 1^2}+frac{1}{3^2 2^2}+frac{1}{3^2 3^3}+dots=frac{1}{9}(frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+dots)= frac{pi^2}{54}$
$frac{1}{2^2}+frac{1}{5^2}+frac{1}{8^2}+dots=frac{1}{2^2 1^2}+frac{1}{3^2?? 1^2}+frac{1}{3^2 ??1^2}$
answered Jan 3 at 16:37
DaleDale
1,1801334
$endgroup$
1
$begingroup$
How does this answer the OP's question?
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
$begingroup$
KM101 deleted his hint... not sure why.
$frac{1}{3^2}+frac{1}{6^2}+frac{1}{9^2}+dots=frac{1}{3^2 1^2}+frac{1}{3^2 2^2}+frac{1}{3^2 3^3}+dots=frac{1}{9}(frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+dots)= frac{pi^2}{54}$
$frac{1}{2^2}+frac{1}{5^2}+frac{1}{8^2}+dots=frac{1}{2^2 1^2}+frac{1}{3^2?? 1^2}+frac{1}{3^2 ??1^2}$
answered Jan 3 at 16:37
DaleDale
1,1801334
$endgroup$
KM101 deleted his hint... not sure why.
$frac{1}{3^2}+frac{1}{6^2}+frac{1}{9^2}+dots=frac{1}{3^2 1^2}+frac{1}{3^2 2^2}+frac{1}{3^2 3^3}+dots=frac{1}{9}(frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+dots)= frac{pi^2}{54}$
$frac{1}{2^2}+frac{1}{5^2}+frac{1}{8^2}+dots=frac{1}{2^2 1^2}+frac{1}{3^2?? 1^2}+frac{1}{3^2 ??1^2}$
answered Jan 3 at 16:37
DaleDale
1,1801334
edited Jan 13 at 18:15
answered Jan 3 at 16:37
DaleDale
1,1801334
answered Jan 3 at 16:37
DaleDale
1,1801334
answered Jan 3 at 16:37
DaleDale
1,1801334
1,1801334
1
$begingroup$
How does this answer the OP's question?
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
1
$begingroup$
How does this answer the OP's question?
$endgroup$
– Mark Viola
Jan 3 at 20:24
1
1
$begingroup$
How does this answer the OP's question?
$endgroup$
– Mark Viola
Jan 3 at 20:24
$begingroup$
How does this answer the OP's question?
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
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$begingroup$
The first sum you mentioned is $frac{1}{9}$ of the initial sum.
$endgroup$
– Peter
Jan 3 at 16:26
$begingroup$
The first sum can be rewritten as$$frac 19+frac 1{36}+cdots=frac 19left[1+frac 14+frac 19+cdotsright]color{blue}{=frac 19zeta(2)}$$
$endgroup$
– Frank W.
Jan 3 at 16:30
$begingroup$
If I may, but what is this problem for? When would you ever need to calculate each sum?
$endgroup$
– Frank W.
Jan 3 at 16:31
$begingroup$
@Peter that means that the other two sums must be different. as if they were the same the total would be $frac{3}{9}$ the initial sum.
$endgroup$
– Dale
Jan 3 at 16:38
$begingroup$
@Dale Peter said the first sum
$endgroup$
– Frank W.
Jan 3 at 16:43