I came through two types of solutions of the series $\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots$ $$\begin{align*} \frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots &=\sum_{n=1}^{\infty}\frac{1}{n+2}\cdot\frac{1}{n!}\\ &=\sum_{n=1}^{\infty}\frac{n+1}{(n+2)!}\\ &=\sum_{n=1}^{\infty}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\ &=\frac{1}{2} \end{align*}$$ $$\begin{align*} \frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots &=\sum_{n=1}^{\infty}\frac{1}{n+2}\cdot\frac{1}{n!}\\ &=\sum_{n=1}^{\infty}\int_{0}^{1}\frac{x^{n+1}}{n!}dx\\ &=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{x^{n+1}}{n!}dx\\ &=\int_{0}^{1}x\sum_{n=1}^{\infty}\frac{x^{n}}{n!}dx\\ &=\int_{0}^{1}x(e^x-1)dx\\ &=\frac{-1}{2} \end{align*}$$ where am I getting wrong please help!
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$\begingroup$ Probably that only a constant is supposed to be taken out...? $\endgroup$– P.K.Dec 28, 2012 at 17:02
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2$\begingroup$ $x(e^x-1)$ is positive for $x\in(0,1)$, so the value of your last integral can't be negative. $\endgroup$– EckhardDec 28, 2012 at 17:03
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$\begingroup$ Actually, $x(e^x-1)$ integrates to $1/2$, not $-1/2$. $\endgroup$– fbgDec 28, 2012 at 17:04
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1 Answer
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You have $$\int^{1}_{0}x(e^{x}-1)dx=[e^{x}(x-1)-\frac{x^{2}}{2}]|^{1}_{0}=\frac{1}{2}$$
answered Dec 28, 2012 at 17:04
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1$\begingroup$ This is a comment, not an answer. The evaluation of the integral is done by the fundamental theorem of calculus:$$ \int_0^1 x \left( \mathrm{e}^x-1\right) \mathrm{d}x = \left. \left(\left(x-1\right) \mathrm{e}^{x} - \frac{x^2}{2} \right)\right|_{x=0}^{x=1} = \frac{1}{2}$$ $\endgroup$– SashaDec 28, 2012 at 17:08
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$\begingroup$ Thanks for your help. It's my mistake. Very sorry to post things like this. $\endgroup$– asimathDec 28, 2012 at 18:05