One Finite Product

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}}

One Finite Product

More Elementary Proof for Euler’s Sine Expansion

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}

More Elementary Proof for Euler’s Sine Expansion

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

The Basel Problem, Double Integral Method II

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 II

The Basel Problem, Double Integral Method I

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. apostol2013Apostol’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}.

The Basel Problem, Double Integral Method I

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}

A Beautiful Convergent Series