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Zeta logarithmic series

Let \zeta denote the zeta function. Prove that

    \[\sum_{k=1}^{\infty} \left( \log\left(1+\frac{1}{k}\right)-\frac{\zeta(k+1)}{k+1} \right)=0\]

Solution

    \begin{align*} \sum_{k=1}^{\infty} \left ( \frac{\zeta(k+1)}{k+1} - \frac{1}{k+1} \right ) &=\sum_{k= 2}^{\infty} \frac{\zeta(k)-1}{k}\\ &=\sum_{k =2}^{\infty} \sum_{n =2}^{\infty} \frac{1}{kn^k}\\ &=\sum_{n=2}^{\infty} \sum_{k= 2}^{\infty} \frac{1}{kn^k}\\ &=-\sum_{n=2}^{\infty} \left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)\\ &=-\sum_{n= 1}^{\infty} \left(\ln\left(1-\frac{1}{n+1}\right)+\frac{1}{n+1}\right) \\ &-\sum_{n =1}^{\infty} \left(\ln\left(\frac{n}{n+1}\right)+\frac{1}{n+1}\right)\\ &=\sum_{n= 1}^{\infty} \left(\ln\left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right) \\ &=\sum_{n= 1}^{\infty} \left(\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\right) \end{align*}

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Mellin transform integral

Evaluate the integral:

    \[\mathcal{J} = \int_0^\infty x \log x e^{-\sqrt{x}} \, \mathrm{d}x\]

Solution

We are evaluating the Mellin transform of the function f(x)=e^{-\sqrt{x}}.

    \begin{align*} \mathcal{M}\left ( f \right ) &= \int_{0}^{\infty} x^{s-1} e^{-\sqrt{x}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\sqrt{x}}{=\! =\! =\! =\! =\!} 2\int_{0}^{\infty} u^{2s-1} e^{-u} \, \mathrm{d}u\\ &= 2 \Gamma\left ( 2s \right ) \end{align*}

where \Gamma is the Euler’s Gamma function. Hence,

    \begin{align*} \int_{0}^{\infty} x \log x e^{-\sqrt{x}} \, \mathrm{d}x &= \mathcal{M}'(f) \bigg|_{s=2} \\ &= \left ( 2 \Gamma(2s) \right )'\bigg|_{s=2}\\ &= 4 \Gamma(2s) \psi^{(0)}(2s) \bigg|_{s=2}\\ &=4 \Gamma(4) \psi^{(0)}(4) \\ &= 4 \cdot 6 \cdot \left ( \frac{11}{6} - \gamma \right ) \\ &=44 - 24 \gamma \end{align*}

where \gamma is the Euler – Mascheroni constant and \psi^{(0)} is the digamma.

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Theta Jacobi limit

Prove that

    \[\lim_{x \rightarrow 1-} \sqrt{1-x} \sum_{n=0}^{\infty} x^{n^2} = \frac{\sqrt{\pi}}{2}\]

Solution

We are working near 1. It follows from the Integral Comparison Test that

    \[\int_0^\infty x^{t^2}\,\mathrm{d}t \leq \sum_{n=0}^\infty x^{n^2} \leq 1 + \int_0^\infty x^{t^2}\,\mathrm{d}t\]

However,

    \begin{align*} \int_0^\infty x^{t^2}\,\mathrm{d}t &= \int_0^\infty \exp\left[-\left(t\sqrt{-\log x}\right)^2\right]\,\mathrm{d}t \\ &= \frac{1}{\sqrt{-\log x}} \int_0^\infty e^{-u^2}\,\mathrm{d}u \\ &= \frac{1}{2} \sqrt{\frac{\pi}{-\log x}} \end{align*}

The result follows from the Sandwich Theorem.

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Double binomial sum

Evaluate the double sum

    \[\mathcal{S} = \sum_{i=0}^m \sum_{j=0}^n \frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}\]

Solution

We sum diagonally , hence:

    \begin{align*} \sum_{i=0}^{m}\sum_{j=0}^{n}\frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}&\overset{i+j=N}{=\! =\! =\! =\!}\sum_{N=0}^{m+n}\sum_{i=0}^m\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &\stackrel{(1)}{=}\sum_{N=0}^{m+n}\sum_{i=0}^{N}\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &=\sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}\\ &\stackrel{(2)}{=} \sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\binom{m+n}{N}\\ &=m+n+1 \end{align*}

(1): For i=m+1,\ldots,N it holds \binom{m}{i}=0.

(2): \displaystyle \sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}=[x^N](1+x)^{m+n}=\binom{m+n}{N}.

Conjecture: Does the following equality

    \[\sum_{a_1=0}^{b_1}\sum_{a_2=0}^{b_2}\cdots\sum_{a_n=0}^{b_n}\frac{\binom{b_1}{a_1}\binom{b_2}{a_2}\cdots\binom{b_n}{a_n}}{\binom{b_1+b_2+\cdots+b_n}{a_1+a_2+\cdots+a_n}}=b_1+b_2+\cdots+b_n+1\]

hold?

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Arctan series

Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \arctan \frac{1}{n^2}\]

Solution

One can argue that the same technique used to evaluate the sum here can be used here as well. Unfortunately, this is not the case as the sum does not telescope. However the technique used here is the way to go.

First of all we note that

    \[\sum_{n=1}^{\infty} \arctan \frac{1}{n^2} = -\arg \prod _{n=1}^{\infty} \left ( 1 - \frac {i} {n^2}\right )\]

However, it is known that

(1)   \begin{equation*}  \sin \pi z =\pi z \prod _{n=1}^{\infty} \left ( 1 - \frac {z^2} {n^2} \right )  \end{equation*}

Let z= \frac {\sqrt {2}}{2} +  \frac {i\sqrt {2}}{2}. We note that z^2=i. Hence,

    \begin{align*} \sum_{n=1}^{\infty} \arctan \frac{1}{n^2} &= -\arg \prod _{n=1}^{\infty} \left ( 1 - \frac {i} {n^2}\right ) \\ &=-\arg \left (\frac {\sin \left ( \frac { \pi \sqrt {2}}{2} + \frac {i \pi \sqrt {2}}{2}\right ) } { \frac {\pi \sqrt {2}}{2} + \frac {i \pi\sqrt {2}}{2}} \right ) \end{align*}

which is now a matter of calculations. The sum is equal to

    \[\sum_{n=1}^{\infty} \arctan \frac{1}{n^2} = \frac{\pi }{4} -\arctan \left ( \cot \frac {\pi \sqrt 2}{2} \tanh \frac {\pi \sqrt 2}{2}\right )\]

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