How to take every third element in a series?

Question

$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$ 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

Answer 1

KM101 deleted his hint... not sure why.

$\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+\dots=\frac{1}{3^2 1^2}+\frac{1}{3^2 2^2}+\frac{1}{3^2 3^3}+\dots=\frac{1}{9}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots)= \frac{\pi^2}{54}$

$\frac{1}{2^2}+\frac{1}{5^2}+\frac{1}{8^2}+\dots=\frac{1}{2^2 1^2}+\frac{1}{3^2?? 1^2}+\frac{1}{3^2 ??1^2}$

Answer 2

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

$$\psi'(1/3)+\psi'(2/3) = 4\pi^2/3$$

Answer 3

As a complement to Mark's answer,

$$\sum_{n\geq 0}\frac{1}{(3n+1)^2}=-\int_{0}^{1}\sum_{n\geq 0} x^{3n}\log(x)\,dx=\int_{0}^{1}\frac{-\log x}{1-x^3}\,dx $$ (and similarly $\sum_{n\geq 0}\frac{1}{(3n+2)^2}$) 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 $\text{Li}_2(a)=\sum_{n\geq 1}\frac{a^n}{n^2}$. 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 $$\text{Re}\,\text{Li}_2(e^{i\theta})=\sum_{n\geq 1}\frac{\cos(n\theta)}{n^2} $$ is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $\text{Im}\,\text{Li}_2(e^{i\theta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function

Answer 4

Polylogarithms

A useful formula that can be applied here is $$ \frac13\sum_{k=0}^2e^{2\pi ijk/3}=[3\mid j]\tag1 $$ So $$ \begin{align}\newcommand{\Li}{\operatorname{Li}} \sum_{j=0}^\infty\frac1{(3j+1)^2} &=\frac13\sum_{k=0}^2\sum_{j=1}^\infty e^{2\pi i(j-1)k/3}\frac1{j^2}\\ &=\frac13\left(\frac{\pi^2}6+e^{-2\pi i/3}\Li_2\left(e^{2\pi i/3}\right)+e^{2\pi i/3}\Li_2\left(e^{-2\pi i/3}\right)\right)\tag2 \end{align} $$ Mathematica gives $1.12173301393634378687$ using
N[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]

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Extended Harmonic Numbers

Another approach is to use the Extended Harmonic Numbers. $$ H(x)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag3 $$ where $$ H'(x)=\sum_{k=1}^\infty\frac1{(k+x)^2}\tag5 $$ Giving $$ \frac19H'\!\left(-\frac23\right)=\sum_{k=0}^\infty\frac1{(3k+1)^2}\tag6 $$ Mathematica gives $1.1217330139363437869$ using
N[1/9HarmonicNumber'[-2/3],20]

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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. $$ \begin{align} \sum_{k=0}^n\frac1{(3k+1)^2} &\sim C-\frac1{3(3n+1)}+\frac1{2(3n+1)^2}-\frac1{2(3n+1)^3}+\frac9{10(3n+1)^5}\\ &-\frac{81}{14(3n+1)^7}+\frac{729}{10(3n+1)^9}-\frac{32805}{22(3n+1)^{11}}\tag7 \end{align} $$ Using $n=100$ in $(7)$, we get $$ \sum_{k=0}^\infty\frac1{(3k+1)^2}=1.1217330139363437868657782\tag8 $$