In this post, I wish to find the value of some finite product. I begin by defining a function $F_n(z)$ as follows and trying to find it’s roots

$\displaystyle F_n(z)=\frac1{2i}\left((1+\frac{iz}n)^n-(1-\frac{iz}n)^n\right)=$ $\displaystyle\frac{(-1)^m}{n^n}$ $\displaystyle z^n+\cdots+$ $\displaystyle1$ $\displaystyle z$

Let $z=n\tan(\theta)$ , then we have

$\displaystyle{1+\frac{iz}n=1+i\tan\theta=1+i\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}(\cos\theta+i\sin\theta)=\sec\theta e^{i\theta}}$

and similarly $1-\frac{iz}n=\sec\theta e^{-i\theta}$ . So we have

$\displaystyle F_n(n\tan\theta)=\frac1{2i}\sec^n\theta(e^{in\theta}-e^{-in\theta})=\sec^n\theta\sin n\theta$

The sine function vanishes at integer multiples of $\pi$ , so it follows that $F_n(n\tan\theta)=0$ where $n\theta=k\pi$ for all integers $k$ , that is, for $\theta=k\pi/n$ for all $k\in\mathbb Z$ . Thus, $F_n(z_k)=0$ for

$\displaystyle z_k=n\tan\frac{k\pi}n=(2m+1)\tan\frac{k\pi}{2m+1}$

where we recall that $n=2m+1$ . Since $\tan\theta$ is strictly increasing on the interval $(-\pi/2,\pi/2)$ , it follows that

$\displaystyle{z_{-m}<z_{-m+1}<...<z_{-1}<z_0<z_1<...<z_{m-1}<z_m}$

moreover, since tangent is an odd function, we have $z_{-k}=-z_k$ for each $k$ . In particular we have found $2m+1=n$ distinct roots of $F_n(z)$ , so as a consequence of the fundamental theorem of algebra, we can write

$\displaystyle F_n(z)=$ $\displaystyle\frac{(-1)^m}{n^n}$ $\displaystyle{z\prod_{k=1}^m(z^2-z_k^2)=\frac{(-1)^m}{n^n}z\prod_{k=1}^m(z^2-n^2\tan^2\frac{k\pi}n)}$

$\displaystyle=\frac{(-1)^m}{n^n}\prod_{k=1}^m(-n^2\tan^2\frac{k\pi}n)z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)$

$\displaystyle=$ $\displaystyle\frac1 n\prod_{k=1}^m\tan^2\frac{k\pi}{n}$ $\displaystyle z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)$

So we have $\frac1{2m+1}\prod_{k=1}^m\tan^2\frac{k\pi}{2m+1}=1$ , and

$\boxed{\displaystyle\prod_{k=1}^m\tan\frac{k\pi}{2m+1}=\sqrt{2m+1}}$

More Elementary Proof for Euler’s Sine Expansion

May 14, 2014December 6, 2020 mathEuler sine expansion, productsLeave a comment

Our goal in this post is to provide an elementary proof for the real version of Euler’s famous Sine expansion.

I first prove that for any $n$ that is a power of $2$ and for any $x\in\mathbb R$ ,

$\displaystyle{\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~(1)$

by using the following more familiar identity

$\displaystyle\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2$

we can easily arrive to the following identity that is valid for $n$ equal to any power of $2$ ,

$\displaystyle\sin(x)=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n$

$\displaystyle=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n$

$\displaystyle=2^{n-1}\sin\frac x n\cos\frac x n\prod_{k=1}^{\frac n 2-1}\left(\sin^2\frac{k\pi}n-\sin^2\frac x n\right)~~~~~~~~~~(2)$

By considering what happens as $x\to0$ in the recent formula, we can obtain that for $n$ a power of $2$

$\displaystyle n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n~~~~~~~~~~(3)$

and then I replace $(3)$ in $(2)$ and prove $(1)$ .

Now I choose $m<\frac n 2-1$ and break up $p_n(x)$ as follows

$\displaystyle p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)~~~~~~~~~~(4)$

Since for $0<\theta<\frac{\pi}2$ , $\sin(\theta)>\frac2{\pi}\theta$ , we have

$\displaystyle\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}$

and if I choose $m$ and $n$ large enough such that $\frac{x^2}{4m^2}<1$ , then

$\displaystyle0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1$

which implies that

$\displaystyle0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1$

Therefore from $(4)$ we have

$\displaystyle p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$

Also we have

$\displaystyle{\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m}$

and once again by $(4)$ , we get

$\displaystyle p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$

To summarize, I have shown that

$\displaystyle{(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~~~~~~(5)$

Now taking $n\to\infty$ in $(5)$ I arrive to

$\displaystyle(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)$

After rearrangement and then taking absolute values, we get

$\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|$

but since we have

$\displaystyle{\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}}$

I can write

$\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}$

Finally since $\lim_{m\to\infty}s_m=0$ , this inequality implies our goal, indeed

$\boxed{\displaystyle\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x}$

One Finite Sum

May 7, 2014November 25, 2020 mathfinite sume, Series, trigonometric identitiesLeave a comment

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,

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

Euler Sum

December 26, 2013November 25, 2020 mathEuler Sums, Leonhard Euler, Series, zeta functionLeave a comment

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:

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

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

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

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

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

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

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

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

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

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

Another Result By the Flajolet-Vardi Theorem

December 7, 2013November 25, 2020 mathflajolet-vardi theorem, Series, zetaLeave a comment

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 II

November 7, 2013November 24, 2020 mathSeries, zeta functionLeave a comment

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

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

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

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

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

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

The Basel Problem, Double Integral Method II

October 28, 2013November 25, 2020 mathBasel problem, Double Integral, Monotone Convergence TheoremLeave a comment

One another way that can evaluate $\zeta(2)$ by double integral is to write $\zeta(2)$ as follows

$\displaystyle\zeta(2)=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$

This is true, since we have

$\displaystyle{\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}}.$

Now integrating by parts yields

$\displaystyle{\int_0^1(-x^{2n}\ln(x))dx=\left[-\frac{x^{2n+1}}{2n+1}\ln(x)\right]_0^1+\int_0^1\frac{x^{2n}}{2n+1}dx=\frac{1}{(2n+1)^2}}$

and by using the monotone convergence theorem, I can write

$\displaystyle{\zeta(2)=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\sum_{n=0}^{\infty}\int_0^1(-x^{2n}\ln(x))dx=\frac{4}{3}\int_0^1\frac{\ln(x)}{x^2-1}dx}$

Hence,

$\displaystyle\zeta(2)=\frac{4}{3}\int_0^1\frac{\ln(x)}{x^2-1}dx$

$\displaystyle=\frac{4}{3}\times\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)$

$\displaystyle=\frac{2}{3}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)$

$\displaystyle=\frac{2}{3}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)$

$\displaystyle=\frac{2}{3}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{2}{3}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du$

$\displaystyle=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\ln(\frac{1}{u^2})du=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du$

$\displaystyle=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du$

$\displaystyle{=\frac{1}{3}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{3}\int_0^{\infty}\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\,dv}$

$\displaystyle{=\frac{1}{3}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\arctan(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{3}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv}$

$\displaystyle{=\frac{\pi}{3}\int_0^{\infty}\frac{1}{1+w^2}dw=\frac{\pi}{3}\left[\arctan(w)\right]_0^{\infty}=\frac{\pi}{3}\times\frac{\pi}{2}=}$ $\displaystyle{\frac{\pi^2}{6}}.$

The Basel Problem, Double Integral Method I

October 21, 2013November 25, 2020 mathBasel problem, Double Integral, Monotone Convergence TheoremLeave a comment

In 1644, the Italian mathematician Pietro Mengoli (1625-1686) posed the question: What’s the value of the sum

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

First time, Leonhard Euler (1707-1783) in 1735 proved that above series converges to $\displaystyle\frac{\pi^2}{6}$ . In this post you can see an easy proof by using double integral that published by Tom M. Apostol in 1983 in Mathematical Intelligencer. Apostol’s Proof: Note that

$\displaystyle{\frac{1}{n^2}}=\displaystyle\int_0^1\displaystyle\int_0^1x^{n-1}y^{n-1}dx\,dy$

and by the monotone convergence theorem we get

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\displaystyle\int_0^1\displaystyle\int_0^1\left(\sum_{n=1}^{\infty}(xy)^{n-1}\right)dx\,dy$

$=\displaystyle\int_0^1\displaystyle\int_0^1\frac{1}{1-xy}dx\,dy$

We change variables in this by putting $(u,v)=\left(\frac{x+y}{2},\frac{y-x}{2}\right)$ , so that $(x,y)=(u-v,u+v)$ . Hence

$\zeta$ $(2)$ $=2\displaystyle\iint_S\frac{du\,dv}{1-u^2+v^2}$

where $S$ is the square with vertices $(0,0), \left(\frac{1}{2},\frac{-1}{2}\right), (1,0)$ and $(\frac{1}{2},\frac{1}{2})$ .

Exploiting the symmetry of the square we get

$\zeta$ $(2)$ $=4\displaystyle\int_0^{\frac{1}{2}}\displaystyle\int_0^u\frac{dv\,du}{1-u^2+v^2}+4\displaystyle\int_{\frac{1}{2}}^1\displaystyle\int_0^{1-u}\frac{dv\,du}{1-u^2+v^2}$

$=4\displaystyle\int_0^{\frac{1}{2}}\frac{\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du+4\displaystyle\int_{\frac{1}{2}}^1\frac{\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$

Now $\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=\sin^{-1}(u)$ , and if $\alpha=\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)$ then $\tan^2(\alpha)=\frac{1-u}{1+u}$ and $\sec^2(\alpha)=\frac{2}{1+u}$ .

It follows that $u=2\cos^2(\alpha)-1=\cos(2\alpha)$ and so $\alpha=\frac{1}{2}\cos^{-1}(u)=\frac{\pi}{4}-\frac{\sin^{-1}(u)}{2}$ . Hence

$\zeta$ $(2)$ $=4\displaystyle\int_0^{\frac{1}{2}}\frac{\sin^{-1}(u)}{\sqrt{1-u^2}}du+4\displaystyle\int_{\frac{1}{2}}^1\frac{\frac{\pi}{4}-\frac{\sin^{-1}(u)}{2}}{\sqrt{1-u^2}}du$

$\displaystyle=\left[2(\sin^{-1}(u))^2\right]_0^{\frac{1}{2}}+\left[\pi\sin^{-1}(u)-(\sin^{-1}(u))^2\right]_{\frac{1}{2}}^1$

$\displaystyle=\frac{\pi^2}{18}+\frac{\pi^2}{2}-\frac{\pi^2}{4}-\frac{\pi^2}{6}+\frac{\pi^2}{36}=$ $\displaystyle\frac{\pi^2}{6}.$

A Beautiful Convergent Series

October 14, 2013November 25, 2020 mathflajolet-vardi theorem, liebniz seriesLeave a comment

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 (Flajolet–Vardi): 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

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

Analysis of Math

Author: math

Solutions to Tom M. Apostol Book on Analytic Number Theory

One Finite Product

More Elementary Proof for Euler’s Sine Expansion

One Finite Sum

Euler Sum

Another Result By the Flajolet-Vardi Theorem

A Beautiful Convergent Series II

The Basel Problem, Double Integral Method II

The Basel Problem, Double Integral Method I

A Beautiful Convergent Series