Strange

Calculus (Integrals, Series)
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Strange

#1

Post by Tolaso J Kos »

Here is something that caught my attention today and I think it is quite strange because someone would expect the series to diverge.

Prove that:

$$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{2n+1} =\frac{\pi}{2}$$
Imagination is much more important than knowledge.
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Strange

#2

Post by Riemann »

Here is a solution:

\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{2n+1} &= \sum_{n=-\infty}^{-1} \frac{(-1)^n}{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}\\
&= \frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{(-1)^n}{1-2n}\\
&= \frac{\pi}{4} - \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\\
&= \frac{\pi}{4} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \\
&= \frac{\pi}{4} + \frac{\pi}{4}\\
&= \frac{\pi}{2}
\end{align*}

since $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}=\frac{\pi}{4}$. It is what we call the Liebniz formula.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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 5 guests