One Finite Sum

In this post I want to find the value of the following finite sum,

\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}

I use the following amazing identity that you can see it’s proof in my previous posts

\displaystyle\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}

By this identity, I can write

                    \displaystyle\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}

                                                           \displaystyle=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^2}{m^2}}

                                                          \displaystyle=m^2\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(x+k\pi+mn\pi)^2}

                                                          \displaystyle=m^2\sum_{n\in\mathbb{Z}}\sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{m^2}{\sin^2(x)}

and this follows that

\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}

Hence,

\boxed{\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}=\lim_{x\to0}\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}=\frac{m^2-1}{3}}

One Finite Sum

Euler Sum

Sums of the form \displaystyle E(p,q)=\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q} where \displaystyle H_n^{(p)}=\sum_{m=1}^n\frac{1}{m^p}, sometimes are called Euler sums. There are several ways to evaluate these  sums. Aabout 240 years ago, Leonhard Euler(1707-1783) in 1775 proved that for q\geq2,

\boxed{\displaystyle{\sum_{n=1}^\infty\frac{H_n}{n^q}=\frac{q+2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\zeta(q-r)\zeta(r+1)}}

In the following you can see an elementary proof of this formula in the three steps:

Step 1:

\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}=\sum_{n=1}^\infty\sum_{m=1}^n\frac{1}{mn^q}

                    \displaystyle=\sum_{m=1}^\infty\sum_{n=m}^\infty\frac{1}{mn^q}

                    \displaystyle=\zeta(q+1)+\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac{1}{mn^q}

                    \displaystyle=\zeta(q+1)+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m(n+m)^q}

                    \displaystyle=\zeta(q+1)+\frac{1}{2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}

Hence,

\boxed{\displaystyle{\sum_{n=1}^\infty\frac{H_n}{n^q}=\zeta(q+1)+\frac{1}{2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}}}

Step 2:

\displaystyle H_n=\sum_{m=1}^n\frac{1}{m}=\sum_{i=0}^{n-1}\sum_{m=1}^\infty\frac{1}{m+i}-\frac{1}{m+i+1}

                              \displaystyle=\sum_{m=1}^\infty\sum_{i=0}^{n-1}\frac{1}{m+i}-\frac{1}{m+i+1}

                              \displaystyle=\sum_{m=1}^\infty\frac{1}{m}-\frac{1}{n+m}

Now by recent formula that is also an alternate definition of the Harmonic numbers, we can write:

\boxed{\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{mn^q}-\frac{1}{(n+m)n^q}}

Step 3:

\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q} \displaystyle=2\sum_{n=1}^\infty\frac{H_n}{n^q} \displaystyle-\sum_{n=1}^\infty\frac{H_n}{n^q}

\displaystyle{=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}-\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{mn^q}-\frac{1}{(n+m)n^q}}

\displaystyle=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^{q-1}n}-\frac{1}{(n+m)n^{q-1}m}

\displaystyle=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac{1}{nm^{q-1}(m-n)}-\frac{1}{mn^{q-1}(m-n)}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{1}{mn^{q-1}(m-n)}-\frac{1}{nm^{q-1}(m-n)}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{m^{q-2}-n^{q-2}}{m^{q-1}n^{q-1}(m-n)}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\left(\sum_{m,n=1}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}-\sum_{\substack{m,n=1\\m=n}}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}\right)

\displaystyle=2\zeta(q+1)+\frac{q-2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m^{q-r}n^{r+1}}

\displaystyle{=\frac{q+2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\zeta(q-r)\zeta(r+1)}.

Euler Sum

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}}.

Another Result By the Flajolet-Vardi Theorem

A Beautiful Convergent Series II

In this post I want to generalize the given formula at my second post as follows

\boxed{\displaystyle\sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1)=\pi\cot\frac{\pi}{m}}

To prove this , I first prove following useful identity

\displaystyle\frac{1}{\sin^2x}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}

PROOF: (Trigonometric Method) Note that

\displaystyle{\frac{1}{\sin^2x}=\frac{1}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\cos^2\frac{x}{2}}\right)=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\sin^2\frac{\pi+x}{2}}\right)}

\displaystyle=\frac{1}{4^2}\left(\frac{1}{\sin^2\frac{x}{2^2}}+\frac{1}{\sin^2\frac{2\pi+x}{2^2}}+\frac{1}{\sin^2\frac{\pi+x}{2^2}}+\frac{1}{\sin^2\frac{3\pi+x}{2^2}}\right)

Repeatedly applying \displaystyle\frac{1}{\sin^2x}=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\sin^2\frac{\pi+x}{2}}\right), we arrive at the following formula:

\displaystyle\frac{1}{\sin^2x}=\frac{1}{4^k}\sum_{n=0}^{2^k-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}

but

\displaystyle{\frac{1}{4^k}\sum_{n=0}^{2^k-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}=\frac{1}{4^k}\sum_{n=-2^{k-1}}^{2^{k-1}-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}=\lim_{k\to\infty}\sum_{n=-k}^k\frac{1}{(x+n\pi)^2}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}}

and now note that

\displaystyle\sum_{n\in\mathbb{Z}}\frac{1}{x+n}=\pi\cot\pi x

Because,

\displaystyle{\cot x=-\int\frac{1}{\sin^2x}dx=-\int\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}dx=\sum_{n\in\mathbb{Z}}\frac{1}{x+n\pi}}

By using this recent identity we can write

\displaystyle\sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1) \displaystyle=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(m-1)^n-1}{m^n}\frac1{k^{n+1}}

                                                                         \displaystyle=\sum_{k=1}^\infty\frac1k\sum_{n=1}^\infty\left(\frac{(m-1)^n}{m^nk^n}-\frac1{m^nk^n}\right)

                                                                         \displaystyle=\sum_{k=1}^\infty\frac1k\left(\frac{\frac{m-1}{mk}}{1-\frac{m-1}{mk}}-\frac{\frac1{mk}}{1-\frac1{mk}}\right)

                                                                         \displaystyle=\sum_{k=1}^\infty\left(\frac1{\frac1m-k}+\frac1{\frac1m+k-1}\right)

                                                                         \displaystyle=\sum_{k\in\mathbb{Z}}\frac1{\frac1m+k} \displaystyle=\pi\cot\frac{\pi}{m}.

A Beautiful Convergent Series II