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I try to evaluate series ${n\over(n+1)!}$ obviously the term $s_n = {n\over (n+1)!} = {1\over (n-1)! (n+1)}$ converges, which has a limit = $0$ when $n\to \infty$ .Then I was stuck and I don't know how to sum this series and whether it converges.

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    $\begingroup$ $$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$$ $\endgroup$ – Did Jun 27 '16 at 20:58
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Hint. One may write $$ \frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac1{n!}-\frac1{(n+1)!} $$ then terms telescope, we get $$ \sum_{n=0}^N\frac{n}{(n+1)!}=1-\frac1{(N+1)!} $$ then we make $N \to \infty$.

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To prove it the sum converges, compare it to something like $\frac 1{n!}$, which you may know converges. To evaluate the sum, define $t_n=\frac 1{(n+1)!}$ Can you sum $(s_n+t_n)$ and $t_n$ and subtract?

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Your logic for convergence is a little faulty. Merely showing that the sequence decreases to $0$ as $n\to \infty$ is not sufficient. To see this, consider the series $\sum \frac{1}{n}$.

Another way to determine convergence: using the ratio test, one may show $\lim_\limits{ n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$ with $a_n={n\over(n+1)!}$

Thus, we write:

$$\lim_\limits{ n\to\infty}\left|\frac{{n+1\over(n+2)!}}{{n\over(n+1)!}}\right|=\lim_\limits{ n\to\infty}\frac{n+1}{(n+2)!}\cdot\frac{(n+1)!}{n}=\lim_\limits{ n\to\infty}\frac{n+1}{n(n+2)}=0<1$$

Thus, the sum converges. For the exact sum, you may use the telescoping method, as described in the other excellent answers.

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Here's an alternate way to find the sum. Recall that $$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}.$$ We can write the above as: $$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1+\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)!} = 1+x\sum_{n=0}^{\infty}\frac{x^n}{(n+1)!},$$ and so: $$\frac{e^x-1}{x} = \sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}.$$ Differentiating both sides gives: $$\frac{xe^x-e^x+1}{x^2} = \sum_{n=1}^{\infty}\frac{nx^{n-1}}{(n+1)!} = \sum_{n=0}^{\infty}\frac{nx^{n-1}}{(n+1)!}.$$ Evaluating both sides at $x=1$ implies: $$\sum_{n=0}^{\infty}\frac{n}{(n+1)!} = 1.$$

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