Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Probably that only a constant is supposed to be taken out...? – Parth Kohli Dec 28 '12 at 17:02
$x(e^x-1)$ is positive for $x\in(0,1)$, so the value of your last integral can't be negative. – Eckhard Dec 28 '12 at 17:03
Actually, $x(e^x-1)$ integrates to $1/2$, not $-1/2$. – fbg Dec 28 '12 at 17:04
up vote 1 down vote accepted

You have $$\int^{1}_{0}x(e^{x}-1)dx=[e^{x}(x-1)-\frac{x^{2}}{2}]|^{1}_{0}=\frac{1}{2}$$

share|cite|improve this answer
This is a comment, not an answer. The evaluation of the integral is done by the fundamental theorem of calculus:$$ \int_0^1 x \left( \mathrm{e}^x-1\right) \mathrm{d}x = \left. \left(\left(x-1\right) \mathrm{e}^{x} - \frac{x^2}{2} \right)\right|_{x=0}^{x=1} = \frac{1}{2}$$ – Sasha Dec 28 '12 at 17:08
I see. Thanks for notice. – Bombyx mori Dec 28 '12 at 17:10
Thanks for your help. It's my mistake. Very sorry to post things like this. – asimath Dec 28 '12 at 18:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.