Does anyone have any idea how to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$
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Start by noticing that $k=0$ gives a divergent series, and that $k=1$ can be dealt with by a telescoping series technique. Then investigate whether a telescoping series can be used for larger $k$. – hardmath Nov 09 '14 at 09:00
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Hint
Notice that
$$\small\frac{1}{n(n+1)(n+2)\cdots(n+k)}=\frac1k\left(\frac{1}{n(n+1)(n+2)\cdots(n+k-1)}-\frac{1}{(n+1)(n+2)\cdots(n+k)}\right)$$ and then telescope.
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Yes by telescoping! Notice that the two terms in the parenthesis are two consecutive terms of a sequence. – Nov 09 '14 at 09:22
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@pavlos, We find $$\left(\frac1n-\frac1{n+1}\right)+\frac12\left(\frac1{n(n+1)}-\frac1{n(n+1)(n+2)}\right)+\frac13\left(\frac1{n(n+1)(n+2)}-\frac1{n(n+1)(n+2)(n+3)}\right)+\cdots$$ right? Then how do you use Telescoping Series? – lab bhattacharjee Nov 09 '14 at 09:54
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1@labbhattacharjee I don't think that's right. $\frac{1} {k}$ is stable. Try writing the general term as Sami does and then write in the parenthesis the partial sum. Don't touch $\frac{1} {k}$. – pavlos Nov 10 '14 at 13:57
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Here is a way to solve this without telescoping. Define $a_n=\displaystyle\frac{k!}{n(n+1)(n+2)\cdots(n+k)}$. The partial fraction expansion of $a_n$ is $\displaystyle \sum_{m=0}^k\frac{(-1)^m{k\choose m}}{n+m}.$ Let $\displaystyle f(x)=\sum_{m=0}^k\frac{(-1)^m{k\choose m}}{n+m}x^{n+m},$ so that $f(0)=0$, $f(1) = a_n$, and $f'(x)=x^{n-1}(1-x)^k$, by the binomial formula. Thus $a_n=f(1)=\int_0^1f'(x)\,dx=\int_0^1 x^{n-1}(1-x)^k\, dx$. Then
$$ \begin{align*} \sum_{n=1}^\infty \frac{k!}{n(n+1)(n+2)\cdots(n+k)} &= \sum_{n=1}^\infty \int_0^1 x^{n-1}(1-x)^k\,dx.\\ &= \int_0^1 \left(\sum_{n=1}^\infty x^{n-1}(1-x)^{k}\right)\,dx\\ &=\int_0^1(1-x)^{k-1}\,dx\\ &=\frac{1}{k}. \end{align*} $$
Interchanging the summation and integration is valid because of monotone convergence.
Jonas Meyer
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