$\sum_{n=1}^{\infty} \frac{n}{(n+1)!}$ converges I am looking for some help with the following question. How do I prove that the series 
$\displaystyle  \sum_{n=1}^{\infty} \frac{n}{(n+1)!}$ converges, and then how would I find its sum?  
I see that this series $a_n$ converges if the sequence of partial sums $s_n$ converges.
So $s_n = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{30}+... + \frac{n}{(n+1)!}$
I see that $\lim\limits_{n\to\infty}\frac{n}{(n+1)!}= 0 $ as $(n+1)!$ approaches infinity faster than $n$
 A: The ratio test shows it converges. Then we may write,
$$\sum_{n=1}^{\infty} \frac{n}{(n+1)!}=\sum_{n=1}^{\infty} \frac{n+1-1}{(n+1)!}$$
$$=\sum_{n=1}^{\infty} \frac{1}{n!}-\sum_{n=1}^{\infty} \frac{1}{(n+1)!}$$
$$=\sum_{n=1}^{\infty} \frac{1}{n!}-\sum_{n=2}^{\infty} \frac{1}{n!}$$
$$=\frac{1}{1!}=1$$
A: Applying the ratio test (disregarding the absolute values as every term in the series is $>0$) we have
\begin{align}
\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}&=\lim\limits_{n\rightarrow\infty}\frac{\frac{n+1}{(n+2)!}}{\frac{n}{(n+1)!}}\\
&=\lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\frac{(n+1)!}{(n+2)!}\\
&=\lim\limits_{n\rightarrow\infty}\frac{n+1}{n(n+2)}\\
&= 0
\end{align}
Since this quantity is smaller than $1$, the series converges.
A: Try using the ratio test:
$\limsup\left|\frac{a_{n+1}}{a_n}\right|=\limsup\left|\frac{\frac{n+1}{(n+2)!}}{\frac{n}{(n+1)!}}\right|=\limsup \left|\frac{n+1}{(n+2)!}\frac{(n+1)!}{n}\right|=\limsup \left|\frac{n+1}{n(n+2)}\right|=\cdots$
It's pretty clear that the denominator is growing faster than the numerator; so we will get $\left|\frac{a_{n+1}}{a_n}\right|<1$.  So this converges.
If you haven't hit analysis yet, instead of $\limsup$ use $\lim\limits_{n \to\infty}.$
A: I propose another solution: $$\frac{n}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}.$$
Now $$\sum_{n=1}^{\infty}\frac{1}{n!}=e-1$$ (for $e$ the summation starts from $n=0$) and $$\sum_{n=1}^{\infty}\frac{1}{(n+1)!}=\sum_{n=2}^{\infty}\frac{1}{n!}=e-1-1=e-2.$$ Because both series converge, its difference also converges. The advantage is we have immediately the sum of our series, which is
$$\sum_{n=1}^{\infty}\frac{n}{(n+1)!}=(e-1)-(e-2)=1.$$
A: In this case you can show that the series converges and find its sum simultaneously by showing that the sequence of partial sums, $S_N$, is
$$S_N=\sum_{n=1}^N{n\over(n+1)!}=1-{1\over(N+1)!}\to1\quad\text{as }N\to\infty$$
The basic idea is contained in Ahmed S. Attaalla's answer:
$${n\over(n+1)!}={(n+1)-1\over(n+1)!}={1\over n!}-{1\over(n+1)!}$$
so that we have the telescoping sum
$$S_N=\left({1\over1!}-{1\over2!} \right)+\left({1\over2!}-{1\over3!} \right)+\cdots+\left({1\over(N-1)!}-{1\over N!} \right)+\left({1\over N!}-{1\over(N+1)!} \right)$$
