How do you find the sum of this alternating series? $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(n+1)!}.$$ 
I found out from my fellow peers at stack exchange see here, that this series converges from the alternating series test. But how do you find the sum? I know if you use wolfram alpha you get: 0.861528, but my question is what steps you use to achieve it?
 A: Here's a familiar trick for summing such series. This is $f(1)$ where
$$f(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)(n+1)!}.$$
Then
$$f'(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(n+1)!}$$
which you can write in closed form. Integrate to get $f(x)$ etc.
A: $$\begin{array}{rcl}
\displaystyle \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(n+1)!}
&=& \displaystyle \sum_{n=0}^\infty \int_0^1 \frac{(-1)^nx^{2n}}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \sum_{n=0}^\infty \int_0^1 \frac{(-x^2)^n}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \sum_{n=0}^\infty \frac{(-x^2)^n}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \sum_{n=0}^\infty\frac{(-x^2)^{n+1}}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \sum_{n=1}^\infty\frac{(-x^2)^n}{n!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \left( \sum_{n=0}^\infty\frac{(-x^2)^n}{n!} - 1\right) \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \left( e^{-x^2} - 1\right) \ \mathrm dx \\
&=& \displaystyle \int_0^1 \left( e^{-x^2} - 1\right) \ \mathrm d\left(\dfrac1x\right) \\
&=& \displaystyle \left(\dfrac{e^{-x^2} - 1}{x}\right)_0^1 - \int_0^1 \dfrac1x \ \mathrm d\left( e^{-x^2} - 1\right) \\
&=& \displaystyle \left(\dfrac1e-1\right) + 2 \int_0^1 e^{-x^2} \ \mathrm dx \\
&=& \displaystyle \left(\dfrac1e-1\right) + \sqrt\pi \cdot \dfrac2{\sqrt\pi} \int_0^1 e^{-x^2} \ \mathrm dx \\
&=& \displaystyle \left(\dfrac1e-1\right) + \sqrt\pi \cdot \operatorname{erf}(1) \\
\end{array}$$
A: From the Taylor development of $e^x$, you draw
$$e^{-x^2}=1+\sum_{n=1}^\infty\frac{(-1)^nx^{2n}}{n!}=1+x^2\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(n+1)!}.$$
Then by integration,
$$\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)(n+1)!}=\int_0^x\frac{e^{-x^2}-1}{x^2}dx.$$
And by parts,
$$\int_0^x\frac{e^{-x^2}-1}{x^2}dx=-\left.\frac{e^{-x^2}-1}x\right|_0^x-2\int_0^xe^{-x^2}dx.$$
The last integral is well known to have no closed-form expression, unless using the error function.
A: 
I know if you use wolfram alpha you get: 0.861528, but my question is what steps you use to achieve it?

Since the given series satisfies the alternating test, then one may use the alternating series remainder,
$$
\left|S\,-\,\sum_{n=0}^N\,(-1)^n\,a_n\right|=\left|\sum_{n=0}^\infty(-1)^n\,a_n\,-\,\sum_{n=0}^N\,(-1)^n\,a_n\right|\le |a_{N+1}|
$$ giving here, with $N=8$,
$$
\left|S\,-\,\sum_{n=0}^8\,\frac{(-1)^n}{(2n+1)(n+1)!}\right|\le \left|\frac{(-1)^9}{(2\times9+1)(9+1)!}\right|<10^{-7}
$$ and the given sum is such that

$$
S=0.8615277\cdots\,\pm 10^{-6}.
$$

