Calculate the series

Complex Analysis
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Calculate the series

#1

Post by Tolaso J Kos »

Evaluate the series:

$$ \mathbf{(a)} \; \sum_{n=-\infty}^{\infty}\frac{1}{n^2+4} \quad \text{and} \quad \mathbf{(b)} \; \sum_{n=-\infty}^{\infty}(-1)^n \frac{1}{n^2+4}$$
Imagination is much more important than knowledge.
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Re: Calculate the series

#2

Post by Tolaso J Kos »

I'll do the first one and give a hint for the second one.

*******************************************************
I'm proving the more general: \( \displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{a}\coth\left ( \pi a \right ) , \; a>0\)

For this purpose I consider the function \( \displaystyle g(z)=\pi\cot \left ( \pi z \right )f(z)=\frac{\pi \cot \left (\pi z \right )}{z^2+a^2} \) which has poles at \(z_1=ai, \; z_2=-ai \). The residue at \(z_1 \) is \( \displaystyle -\frac{\pi\coth(\pi a)}{2a} \) and similarly the residue at \(z_2 \) is equal to \( \displaystyle -\frac{\pi\coth(\pi a)}{2a} \).

Hence:
$$\sum_{n=-\infty}^{\infty}\frac{1}{n^2+a^2}=-\sum_{residues}=\frac{\pi\coth(\pi a)}{a}$$

It is known that \( \cot(ai)=-i\coth(a) \)

For the second sum consider the function \( \displaystyle g(z)=\pi \csc (\pi z)f(z) \) , evaluate the residues at all poles. The series will then be equal to the minus sum of the residues.
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 9 guests