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?

$$\begin{array}{rcl}
\displaystyle \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(n+1)!}
&=& \displaystyle \sum_{n=0}^\infty \int_0^1 \frac{(-1)^nx^{2n}}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \sum_{n=0}^\infty \int_0^1 \frac{(-x^2)^n}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \sum_{n=0}^\infty \frac{(-x^2)^n}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \sum_{n=0}^\infty\frac{(-x^2)^{n+1}}{(n+1)!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \sum_{n=1}^\infty\frac{(-x^2)^n}{n!} \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \left( \sum_{n=0}^\infty\frac{(-x^2)^n}{n!} - 1\right) \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{-x^2} \left( e^{-x^2} - 1\right) \ \mathrm dx \\
&=& \displaystyle \int_0^1 \left( e^{-x^2} - 1\right) \ \mathrm d\left(\dfrac1x\right) \\
&=& \displaystyle \left(\dfrac{e^{-x^2} - 1}{x}\right)_0^1 - \int_0^1 \dfrac1x \ \mathrm d\left( e^{-x^2} - 1\right) \\
&=& \displaystyle \left(\dfrac1e-1\right) + 2 \int_0^1 e^{-x^2} \ \mathrm dx \\
&=& \displaystyle \left(\dfrac1e-1\right) + \sqrt\pi \cdot \dfrac2{\sqrt\pi} \int_0^1 e^{-x^2} \ \mathrm dx \\
&=& \displaystyle \left(\dfrac1e-1\right) + \sqrt\pi \cdot \operatorname{erf}(1) \\
\end{array}$$
