Need proof for the following infinite series !!

Prove that

1n+a=πcot(πa). \displaystyle \sum_{-\infty}^{\infty} \dfrac1{n+a} = \pi \cot (\pi a ) .

#Calculus

Note by A Former Brilliant Member
4 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

If you ask me how to do this by complex analysis , there is a much simpler answer.

i am using the summation theorem, first take a look here if necessary to see how all this happens. Let the function be f(z)=1z+a\displaystyle f(z)=\frac{1}{z+a} . It has a simple pole at z=az=-a and by summation theroem we need to evaluate the residue of πcot(πz)f(z)\pi\cot (\pi z)f(z) at z=az=-a

The residue is therefore R=limza(z+a)πcotπzz+a=πcotπa\displaystyle R=\lim_{z\to -a}(z+a) \frac{\pi\cot\pi z}{z+a}=-\pi\cot\pi a

The answer is therefore the negative of the residue ,

1n+a=πcotπa\displaystyle \sum_{-\infty}^\infty \frac{1}{n+a}=\pi\cot \pi a

The series is actually the representation of ψ(a)\psi(a) where ψ(.)\psi(.) is the Digamma Function.

Aditya Narayan Sharma - 4 years, 6 months ago

Log in to reply

Wow. This is exactly what I was looking for. Thank you so much for such a wonderful reply.

A Former Brilliant Member - 4 years, 6 months ago

Log in to reply

(Assuming you know complex analysis,) do you see why complex analysis is the natural way to approach this problem?

When asking about "what have you tried", part of it is also to understand how much background you have, in order to provide a suitable reply. Your solution didn't give me great confidence in your understanding of calculus/anlysis, which is why I was unable to suggest taking a "high power" approach instead of a "crude summation like Riemann sum".

In future, by providing this context to others (e.g. I got this problem when working through Serge Lang's complex analysis book on this chapter), that will give you better replies.

Calvin Lin Staff - 4 years, 6 months ago

What have you tried, where did you get stuck?

Calvin Lin Staff - 4 years, 6 months ago

Log in to reply

Thank you so much for the response. I don't know how to deal with it. I know about General Harmonic series. But i am not getting any idea about how to solve a summation from "negative infinity to positive infinity" which also involves a constant "a" .

A Former Brilliant Member - 4 years, 6 months ago
×

Problem Loading...

Note Loading...

Set Loading...