Strange
- Tolaso J Kos
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Strange
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:
$$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{2n+1} =\frac{\pi}{2}$$
Prove that:
$$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{2n+1} =\frac{\pi}{2}$$
Imagination is much more important than knowledge.
Re: Strange
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.
\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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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