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Evaluate the series:

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

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    @did: I just made it more beautiful, more concise.2012-09-02

2 Answers 2

20

Partial fraction decomposition gives

$\frac{1}{k(k+1)^2}=\left(\frac{1}{k}-\frac{1}{k+1}\right)\frac{1}{k+1}=\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}$

Hence this series is

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

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

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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    +1 nice, and $\sum 1/(k\times k!)=\text{Ei}(x) -\gamma$.2012-06-19
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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).2012-09-02