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I want to find the sum of $\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$.

I have tried to turn it into a power series for a known function, with no luck. I also tried to write it as $\sum_{n=0}^{\infty} \frac{(x)^n}{2n+1}$.

$\frac{1}{2\sqrt{x}}\sum_{n=0}^{\infty} \frac{(x)^{n+\frac{1}{2}}}{n+\frac{1}{2}}$

And then differentiated, but that didn't make it any easier (and the {\sqrt{x}} is not real for $x=-1$).

Hints are appreciated.

amWhy
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Per
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1 Answers1

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What you are describing is the Leibniz formula for $\pi$:

\begin{equation} S = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{\pi}{4} \end{equation}

One very simple proof is to consider the Maclaurin series for $\arctan x$:

\begin{equation} \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots \end{equation}

And setting $x = 1$ yields the above sum ($S$) as equal to $\arctan(1) = \frac{π}{4}$. However, if you are more interested in the why than the how behind this sum, I would suggest checking out the awesome video by 3B1B: https://www.youtube.com/watch?v=NaL_Cb42WyY

Martin Westin
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  • Thanks, that helped me out.
    I also want to find $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$. Using the series of arctan(x) and differentiating, i get $\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)^2} = \int_{0}^{x} \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{2n+1} = \int_{0}^{x} \frac{arctan(x)}{x}$, which does not seem helpful. Any hints on this problem?     – Per Nov 16 '23 at 21:31
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    The sum $\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ is known as Catalan's constant: https://en.wikipedia.org/wiki/Catalan%27s_constant – Minus One-Twelfth Nov 16 '23 at 21:45