Sum of the series $\sum {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}$ 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 $\frac 12$.
 A: $\dfrac{\sin x}{2x}-\dfrac{\cos x}2=\dfrac1{2x}\sum\limits_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}-\dfrac12\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}$
$=\dfrac12\sum\limits_{n=0}^\infty(-1)^nx^{2n}\left(\dfrac{1}{(2n+1)!}-\dfrac1{(2n)!}\right)$
$=\sum\limits_{n=0}^\infty\dfrac12(-1)^{n+1}x^{2n}\left(\dfrac{2n+1}{(2n+1)!}-\dfrac1{(2n+1)!}\right)$
$=\sum\limits_{n=0}^\infty(-1)^{n+1}x^{2n}\dfrac n{(2n+1)!}$.
Now take $x=\pi$.
A: $\displaystyle \sum_{n=0}^\infty \textstyle{(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}=\displaystyle {-1\over 2\pi}\sum_{n=0}^\infty \textstyle {(-1)^n2n\over (2n+1)!}\cdot \pi^{2n+1}$
$\pi \cos \pi-\sin \pi=\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n(2n+1) \over (2n+1)!}\cdot \pi^{2n+1}-\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n \over (2n+1)!}\cdot \pi^{2n+1}$
$=\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^n2n\over (2n+1)!}\cdot \pi^{2n+1}$
Hence $\displaystyle \sum_{n=0}^\infty \textstyle {(-1)^{n+1}n\over (2n+1)!}\cdot \pi^{2n}={-1\over 2\pi}(\pi \cos \pi-\sin \pi) =\frac 12$
A: 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: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
