# Another Result By the Flajolet-Vardi Theorem

By using the Flajolet-Vardi theorem we can find the value of the another amazing convergent series. Indeed, following series

$\displaystyle\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$

I first recall the Flajolet-Vardi theorem that you can find it’s proof in my second post:

Flajolet-Vardi Theorem:

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

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

This theorem shows that $\displaystyle\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}=\sum_{n=1}^{\infty}\frac{1}{kn(kn-1)}$, because if we let $f(z)=\displaystyle\sum_{n=2}^{\infty}\frac{z^n}{k^n}$, then $f(z)=\frac{z^2}{k(k-z)}$ and by this theorem

$\displaystyle{\sum_{n=2}^\infty\frac{\zeta(n)}{k^n}=\sum_{n=1}^\infty f\left(\frac{1}{n}\right)=\sum_{n=1}^\infty\frac{1}{kn(kn-1)}}$

Therefore now we must find the value of $\displaystyle\sum_{n=1}^\infty\frac{1}{kn(kn-1)}$. we use the Taylor expansion of $\log(1-x)$ and the fact that the sum $\displaystyle\sum_{\alpha^k=1}\alpha^n$ is $k$ if $k$ divides $n$ and $0$ otherwise.

There are some $\log$‘s of complex numbers. Those numbers have always non-negative real part, for the Argument we take the angle between $\displaystyle\frac{-\pi}{2}$ and $\displaystyle\frac{\pi}{2}$, so that it fits with the power series for $\log(1-x)$.

$\displaystyle{\sum_{n=1}^\infty \frac{x^{kn}}{kn}=\frac{-\log(1-x^k)}{k}=\frac{-1}{k}\sum_{\alpha^k=1}\log(1-\alpha x)}$

$\displaystyle\sum_{\alpha^k=1}\sum_{m=1}^\infty\frac{\alpha(\alpha x)^m}{m}=k\sum_{n=1}^{\infty}\frac{x^{kn-1}}{(kn-1)}$

but also

$\displaystyle{\sum_{\alpha^k=1}\sum_{m=1}^\infty\frac{\alpha(\alpha x)^m}{m}=-\sum_{\alpha^k=1}\alpha\log(1-\alpha x)}.$

We thus have

$\displaystyle{\sum_{n=1}^\infty\frac{x^{kn}}{kn(kn-1)}=\sum_n x^{kn}\left(\frac{1}{kn-1}-\frac{1}{kn}\right)}$

$\displaystyle{=\frac{1}{k}\sum_{\alpha^k=1}(1-x\alpha)\log(1-\alpha x)}.$

We have to take the limit $x\to 1$. The $\alpha=1$ term disappears, so we get

$\displaystyle{\sum_{n=1}^\infty\frac{1}{kn(kn-1)}=\frac{1}{k}\sum_{\alpha^k=1,\alpha\neq1}(1-\alpha)\log(1-\alpha),\alpha=e^{\frac{2\pi im}{k}}}$

$\displaystyle{=\frac{1}{k}\sum_{m=1}^{k-1}\left(1-\cos\frac{2\pi m}{k}-i\sin\frac{2\pi m}{k}\right)\left(\log\left(2\sin\frac{\pi m}{k}\right)+\pi i\left(\frac{m}{k}-\frac{1}{2}\right)\right)}$

$\displaystyle{=\frac{1}{k}\sum_{m=1}^{k-1}\left[\left(1-\cos\frac{2\pi m}{k}\right)\log\left(2\sin\frac{\pi m}{k}\right)+\frac{(2m-k)\pi}{2k}\sin\frac{2\pi m}{k}\right]}.$

Examples:

For $k=2$ we have

$\displaystyle{\sum_{n=2}^\infty\frac{\zeta(n)}{2^n}=\sum_{n=1}^\infty\frac{1}{2n(2n-1)}=\frac{1}{2}\left[(1-\cos\pi)\log\left(2\sin\frac{\pi}{2}\right)\right]=\log(2)}.$

and also for $k=3$,

$\displaystyle\sum_{n=2}^\infty\frac{\zeta(n)}{3^n}=\sum_{n=1}^\infty\frac{1}{3n(3n-1)}$

$\displaystyle=\frac{1}{3}\left[\left(1-\cos\frac{2\pi}{3}\right)\log\left(2\sin\frac{\pi}{3}\right)-\frac{\pi}{6}\sin\frac{2\pi}{3}+\left(1-\cos\frac{4\pi}{3}\right)\log\left(2\sin\frac{2\pi}{3}\right)+\frac{\pi}{6}\sin\frac{4\pi}{3}\right]$

$\displaystyle=\frac{1}{2}\log(3)-\frac{\pi}{6\sqrt{3}}.$

# 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

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