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$$1 - \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} - \frac{1}{7 \cdot 4!}+\cdots$$

I am new to math over flow, and I do not know how to format the math, sorry! Also, what should this converge to?

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  • $\begingroup$ You could try comparison with the exponential series $\sum_{n=0}^\infty 1/n!$. $\endgroup$ – Lord Shark the Unknown Apr 17 '17 at 5:20
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The series is given by $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(n+1)!}.$$ Convergence follows from the Alternating Series Test. Is this test known to you?

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For the convergence, ΘΣΦGenSan gave the tool to use.

For the sum, it is (at least to me) slightly tricky.

Consider $$y=\sum_{n=0}^\infty \frac{(-1)^n x^{2 n+1}}{(2 n+1) (n+1)!}$$ and differentiate $$y'=\sum_{n=0}^\infty \frac{(-1)^n x^{2 n}}{(n+1)!}=\frac{1}{x^2}-\frac{e^{-x^2}}{x^2}$$ Now, integrate to get $$y=-\frac{1}{x}+\frac{e^{-x^2}}{x}+\sqrt{\pi }\,\text{erf}(x)$$ where appears the error function.

Make $x=1$ and get $$-1+\frac{1}{e}+\sqrt{\pi }\, \text{erf}(1)\approx 0.861528$$ If you compute the partial sums $$S_p=\sum_{n=0}^p \frac{(-1)^n x^{2 n+1}}{(2 n+1) (n+1)!}$$ you should notice that this value (six significant figures) is already obtained for $p=7$.

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$$1 - \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} - \frac{1}{7 \cdot 4!}+\cdots$$ A standard theorem says this converges if the corresponding series of absolute values converges: $$1 + \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} + \frac{1}{7 \cdot 4!}+\cdots$$ The "comparison test" says this converges if the following converges: $$1 + \frac 1 {2!} + \frac 1 {3!} + \frac 1 {4!}+\cdots$$ And that converges by a ratio test: $$ \frac{1/(n+1)!}{1/n!} = \frac 1 {n+1} \to \text{ as } n\to\infty. $$

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