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Evaluate the sum of the series: $\displaystyle \sum_{n=0}^\infty {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$

Ratio test confirms convergence of the series. But what's the exact value of the series ? Computer programming approximates it somewhere near $\frac 12$.

Sam
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3 Answers3

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$\dfrac{\sin x}{2x}-\dfrac{\cos x}2=\dfrac1{2x}\sum\limits_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}-\dfrac12\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}$

$=\dfrac12\sum\limits_{n=0}^\infty(-1)^nx^{2n}\left(\dfrac{1}{(2n+1)!}-\dfrac1{(2n)!}\right)$

$=\sum\limits_{n=0}^\infty\dfrac12(-1)^{n+1}x^{2n}\left(\dfrac{2n+1}{(2n+1)!}-\dfrac1{(2n+1)!}\right)$

$=\sum\limits_{n=0}^\infty(-1)^{n+1}x^{2n}\dfrac n{(2n+1)!}$.

Now take $x=\pi$.

J. W. Tanner
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$\displaystyle \sum_{n=0}^\infty \textstyle{(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}=\displaystyle {-1\over 2\pi}\sum_{n=0}^\infty \textstyle {(-1)^n2n\over (2n+1)!}\cdot \pi^{2n+1}$ $\pi \cos \pi-\sin \pi=\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n(2n+1) \over (2n+1)!}\cdot \pi^{2n+1}-\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n \over (2n+1)!}\cdot \pi^{2n+1}$ $=\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n2n\over (2n+1)!}\cdot \pi^{2n+1}$

Hence $\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}={-1\over 2\pi}(\pi \cos \pi-\sin \pi) =\frac 12$

Saikai Prime
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The summation with $x=i\pi$

$$S=\dfrac i\pi\sum_{n=0}^\infty\dfrac{nx^{2n+1}}{(2n+1)! }$$

Writing $2n=2n+1-1, $ $$S=\dfrac i{2\pi}\left(x\sum\dfrac{x^{2n}}{(2n)!}-\sum\dfrac{x^{2n+1}}{(2n+1)!}\right)$$

The first infinite sum $=\dfrac{e^x+e^{-x}}2$

and the second $=\dfrac{e^x-e^{-x}}2$

Use How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

lab bhattacharjee
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