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Evaluate the series:

$$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$$

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did you try out partial fraction ?? – Theorem Jun 19 '12 at 7:16
@Theorem: i suppose that this is the only thing you can do here. I wonder what is the best splitting in order to easily get the answer. Yes, i have some ideas about that, but they didn't work as expected. – user 1618033 Jun 19 '12 at 7:19
Why did you modify the question? – Did Sep 2 '12 at 11:45
@did: I just made it more beautiful, more concise. – user 1618033 Sep 2 '12 at 11:52
up vote 20 down vote accepted

Partial fraction decomposition gives


Hence this series is


$$=\left(\sum_{k=1}^\infty\frac{1}{k \cdot k!}\right)-\left(\sum_{k=1}^\infty\frac{1}{(k+1)!}\right)-\left(\sum_{r=2}^\infty\frac{1}{r\cdot r!}\right).$$

Notice how in the third sum we set $r=k+1$ so $(k+1)^2k!=(k+1)\cdot(k+1)!=r\cdot r!$. The middle term is clearly $e-2$, and the difference between the outside series is $\frac{1}{1\cdot 1!}$, hence we obtain $3-e$.

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It's very useful to have $$\frac{1}{k(k+a)}=\frac{1}{a} \left(\frac{1}{k}-\frac{1}{k+a}\right) ,\quad \frac{1}{k^2-a^2}=\frac{1}{2a}\left(\frac{1}{k-a}-\frac{1}{k+a}\right)$$ memorized. I used the special case $a=1$. – anon Jun 19 '12 at 7:23
that's right! Especially at these series. – user 1618033 Jun 19 '12 at 7:27
I like these witty solutions....+1, of course. – DonAntonio Jun 19 '12 at 10:22
+1 nice, and $\sum 1/(k\times k!)=\text{Ei}(x) -\gamma$. – draks ... Jun 19 '12 at 11:05

Try it with factor $x^k$. $$\begin{align} f(x) &=\sum_{k=1}^{\infty}\frac{x^k}{k(k+1)^2k!} \\ f'(x) &=\sum_{k=1}^{\infty}\frac{x^{k-1}}{(k+1)^2k!} \\ x^2f'(x) &=\sum_{k=1}^{\infty}\frac{x^{k+1}}{(k+1)^2k!} \\ \big(x^2f'(x)\big)' &=\sum_{k=1}^{\infty}\frac{x^{k}}{(k+1)k!} \\ x\big(x^2f'(x)\big)' &=\sum_{k=1}^{\infty}\frac{x^{k+1}}{(k+1)k!} \\ \Big(x\big(x^2f'(x)\big)'\Big)' &=\sum_{k=1}^{\infty}\frac{x^{k}}{k!} = e^x-1 . \end{align}$$ Now solve a differential equation. Then plug in $x=1$.

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beautiful (+1). – user 1618033 Sep 2 '12 at 13:31

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