Home » Posts tagged 'Series'

Tag Archives: Series

Series of Bessel function

October 10, 2019 / Leave a comment

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

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

Solution

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*}

Read more

Zeta logarithmic series

September 15, 2019 / Leave a comment

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*}

Read more

Arctan series

September 2, 2019 / Leave a comment

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

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

Read more

A telescoping arctan series

September 1, 2019 / 2 Comments on A telescoping arctan series

Evaluate the sum

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

Solution

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

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

Read more

An arctan series

September 1, 2019 / Leave a comment

Prove that

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

Solution

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.

Read more

Post navigation

← Older posts

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network