# A Beautiful Convergent Series

In this post I want to find the value of the sum

$\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \zeta \left(n+1 \right)$

Note that for $s>1$, $\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}$.

I use following formula that is called Gregory-Leibniz-Madhava’s series

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n-3}-\frac{1}{4n-1}=\frac{\pi}{4}$

If we define $f$ function as follows:

$\displaystyle{f\left(z\right)=\frac{z}{4-3z}-\frac{z}{4-z}}$

then

$\displaystyle{f(z)=\frac{\frac{z}{4}}{1-\frac{3z}{4}}-\frac{\frac{z}{4}}{1-\frac{z}{4}}=\frac{z}{4}\sum_{n=0}^{\infty}\left(\frac{3z}{4}\right)^n-\left(\frac{z}{4}\right)^n=\frac{z}{4}\sum_{n=1}^{\infty}\left(\frac{3z}{4}\right)^n-\left(\frac{z}{4}\right)^n}$

$\displaystyle{=\frac{z}{4}\sum_{n=2}^{\infty}\left(\frac{3z}{4}\right)^{n-1}-\left(\frac{z}{4}\right)^{n-1}=\frac{1}{4}\sum_{n=2}^{\infty}\left[\left(\frac{3}{4}\right)^{n-1}-\left(\frac{1}{4}\right)^{n-1}\right]z^n=\frac{1}{4}\sum_{n=2}^{\infty}\frac{3^{n-1}-1}{4^{n-1}}z^n}.$

Now I use this theorem

THEOREM (FlajoletVardi): If $f\left(z \right)=\displaystyle\sum_{n=2}^{\infty}a_{n}z^n$ and $\displaystyle\sum_{n=2}^{\infty}|a_n|$ converges then,

$\displaystyle\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\displaystyle\sum_{n=2}^{\infty}a_n$$\zeta$$\left(n\right)$.

PROOF:

Because $\displaystyle\sum_{m=2}^{\infty}|a_m|<\infty$,

$\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=2}^{\infty}|a_m|\frac{1}{n^m}\leq\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=2}^{\infty}|a_m|\frac{1}{n^2}<\infty$

Hence, by Cauchy’s double series theorem, we can switch the order of summation:

$\displaystyle{\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}a_m\frac{1}{n^m}=\sum_{m=2}^{\infty}a_m\sum_{n=1}^{\infty}\frac{1}{n^m}=\sum_{n=2}^{\infty}a_n\zeta(n)}$

This theorem implies that

$\displaystyle{\frac{\pi}{4}=\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\frac{1}{4}\sum_{n=2}^{\infty}\frac{3^{n-1}-1}{4^{n-1}}\zeta(n)=\frac{1}{4}\sum_{n=1}^{\infty}\frac{3^{n}-1}{4^{n}}\zeta(n+1)}$

and

$\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \zeta \left(n+1 \right)=\pi}$