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Series of Bessel function

Let J_n denote the Bessel function of the first kind. Prove that

    \[\sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 =1\]


The Jacobi – Anger expansion tells us that

(1)   \begin{equation*} e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}  \end{equation*}

Hence by Parseval’s Theorem it follows that

    \begin{align*} \sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,\mathrm{d} \theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d} \theta \\ &=1 \end{align*}

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Zeta logarithmic series

Let \zeta denote the zeta function. Prove that

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


    \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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Arctan series

Evaluate the series

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


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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A telescoping arctan series

Evaluate the sum

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


First of all note that:

    \begin{align*} \sum_{n=1}^N \arctan \frac{2}{n^2} &=\sum_{n=1}^N \arctan (n+1)-\arctan(n-1) \\ &=-\frac{\pi}{4}+\arctan N+\arctan(N+1) \end{align*}

Hence letting N \rightarrow +\infty we have that

    \[\sum_{n=1}^{\infty} \arctan \frac{2}{n^2} = \frac{3 \pi}{4}\]

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An arctan series

Prove that

    \[\sum_{n=1}^{\infty}\arctan \left(\frac{10n}{(3n^2+2)(9n^2-1)}\right) =\ln{3}-\frac{\pi}{4}\]


The key ingredient is the observation \arctan x=\arg(1+ix). Then we note that

    \begin{align*} 1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)} &= \frac{\left ( n-i \right )\left ( 3n-\left ( 1-i \right ) \right )\left ( 3n+i \right )\left ( 3n+ \left ( 1+i \right ) \right )}{\left ( 3n-1 \right )\left ( 3n+1 \right )\left ( 3n^2+2 \right )} \\ &= \frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}} \end{align*}

Using the arg’s property we get that

    \begin{align*} \arctan \left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right) &=\arctan\left(\frac1{3n-1}\right)+\arctan \frac{1}{3n}+\\ &\quad  + \arctan\left(\frac1{3n+1}\right)-\arctan \frac{1}{n} \end{align*}

Hence the initial sum telescopes;

    \begin{align*} \sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right) &=\lim_{N \rightarrow +\infty}\sum_{n=1}^N \bigg[ \arctan\left(\frac1{3n-1}\right)+\\ & \quad  + \arctan \frac{1}{3n}+ \arctan \left(\frac{1}{3n+1} \right)- \\ &\quad \quad \quad - \arctan \frac{1}{n} \bigg] \\ &=\lim_{N \rightarrow +\infty}\sum_{n=N+1}^{3N+1}\arctan \frac{1}{n} - \arctan 1\\ &=\lim_{N \rightarrow +\infty} \sum_{n=N+1}^{3N+1}\left[\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right] - \\ &\quad \quad \quad \quad -\arctan 1\\ &=\ln 3-\frac{\pi}{4} \end{align*}

since \displaystyle \sum_{n=N+1}^{3N+1} \frac{1}{n} \sim \ln \left ( \frac{3N+1}{N+1} \right ) which explains why \ln 3 pops up.

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