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$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(n+1)!}.$$ 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?

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  • $\begingroup$ $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac{(-1)^n}{(2 n + 1) (n + 1)!} = \sqrt \pi \operatorname{erf} \left( 1 \right) - 1 + \dfrac 1e$, so unless you are really interested, it would not be very useful. $\endgroup$ – DHMO Apr 17 '17 at 7:44
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    $\begingroup$ This particular sum admits the closed form $$\frac{e\sqrt{\pi} \text{erf}(1) + 1 - e}{e} \approx 0.861528$$ I'm sure someone on this site can prove this. Note that the decimal approximation is just that: an approximation. However, note that, in general, if a series converges, it does not necessarily have a nice expression for it. But, computers may still give numerical approximations. $\endgroup$ – MathematicsStudent1122 Apr 17 '17 at 7:45
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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.

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  • $\begingroup$ In fact the closed form contains $e^{-x^2}$ as shown by my answer. It is not as simple as that. $\endgroup$ – DHMO Apr 17 '17 at 7:54
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$$\begin{array}{rcl} \displaystyle \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(n+1)!} &=& \displaystyle \sum_{n=0}^\infty \int_0^1 \frac{(-1)^nx^{2n}}{(n+1)!} \ \mathrm dx \\ &=& \displaystyle \sum_{n=0}^\infty \int_0^1 \frac{(-x^2)^n}{(n+1)!} \ \mathrm dx \\ &=& \displaystyle \int_0^1 \sum_{n=0}^\infty \frac{(-x^2)^n}{(n+1)!} \ \mathrm dx \\ &=& \displaystyle \int_0^1 \dfrac1{-x^2} \sum_{n=0}^\infty\frac{(-x^2)^{n+1}}{(n+1)!} \ \mathrm dx \\ &=& \displaystyle \int_0^1 \dfrac1{-x^2} \sum_{n=1}^\infty\frac{(-x^2)^n}{n!} \ \mathrm dx \\ &=& \displaystyle \int_0^1 \dfrac1{-x^2} \left( \sum_{n=0}^\infty\frac{(-x^2)^n}{n!} - 1\right) \ \mathrm dx \\ &=& \displaystyle \int_0^1 \dfrac1{-x^2} \left( e^{-x^2} - 1\right) \ \mathrm dx \\ &=& \displaystyle \int_0^1 \left( e^{-x^2} - 1\right) \ \mathrm d\left(\dfrac1x\right) \\ &=& \displaystyle \left(\dfrac{e^{-x^2} - 1}{x}\right)_0^1 - \int_0^1 \dfrac1x \ \mathrm d\left( e^{-x^2} - 1\right) \\ &=& \displaystyle \left(\dfrac1e-1\right) + 2 \int_0^1 e^{-x^2} \ \mathrm dx \\ &=& \displaystyle \left(\dfrac1e-1\right) + \sqrt\pi \cdot \dfrac2{\sqrt\pi} \int_0^1 e^{-x^2} \ \mathrm dx \\ &=& \displaystyle \left(\dfrac1e-1\right) + \sqrt\pi \cdot \operatorname{erf}(1) \\ \end{array}$$

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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.

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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

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

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