4

show that and determine the closed form

$$\sum_{n=1}^{\infty}{m!\over n\cdot n(n+1)(n+2)\cdots(n+m)}=\sum_{n=1}^{\infty}{{1\over n} }\sum_{k=0}^{m}(-1)^k{m\choose k}{1\over (n+k)}$$

$m\ge0$

My try:

I was observing the favourite Euler's sum

$$\zeta(2)=\sum_{n=1}^{\infty}{1\over n^2}$$

I found the following patterns. I can't find the closed form.

$$\sum_{n=1}^{\infty}{0!\over n\cdot n}=\sum_{n=1}^{\infty}{{1\over n}\left({1\over n}\right)}$$

$$\sum_{n=1}^{\infty}{1!\over n\cdot n(n+1)}=\sum_{n=1}^{\infty}{{1\over n}\left({1\over n}-{1\over n+1}\right)}$$

$$\sum_{n=1}^{\infty}{2!\over n\cdot n(n+1)(n+2)}=\sum_{n=1}^{\infty}{{1\over n}\left({1\over n}-{2\over n+1}+{1\over n+2}\right)}$$

$$\sum_{n=1}^{\infty}{3!\over n\cdot n(n+1)(n+2)(n+3)}=\sum_{n=1}^{\infty}{{1\over n}\left({1\over n}-{3\over n+1}+{3\over n+2}-{1\over n+3}\right)}$$

gymbvghjkgkjkhgfkl
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  • I got an answer but not a closed form $\zeta(2)-1-{1\over 4}-{1\over9}\cdots-{1\over m^2}$. This is the best I can do for now. – gymbvghjkgkjkhgfkl Dec 31 '16 at 20:00
  • $$ \zeta(2)-\sum_{n=1}^{m}\frac{1}{n^2} = \sum_{n=m+1}^{\infty}\frac{1}{n^2} = \sum_{n=0}^{\infty}\frac{1}{(m+1+n)^2} = \zeta(2,m+1) $$ – Hazem Orabi Dec 31 '16 at 20:47
  • I think the series is a series representation of Hurwitz zeta function: $$ \zeta(s,q) = \frac{1}{s-1}\sum_{n=0}^{\infty}\frac{1}{n+1},\sum_{k=0}^{n}(-1)^k{n\choose k}(q+k)^{1-s} \ \Rightarrow \zeta(2,m+1) = \sum_{n=1}^{\infty}\frac1n,\sum_{k=1}^{\color{red}{n}}(-1)^k{n\choose k}\frac{1}{\color{red}{m}+k} $$ – Hazem Orabi Dec 31 '16 at 21:08

1 Answers1

5

We can do a partial fraction expansion of the summand as follows. Start by writing

$$\frac{m!}{n(n+1)\cdots(n+m)} = \sum_{i=0}^m \frac{A_i}{n+i}$$

This identity holds for all values of $n$ not just integers. For any integer $k\in\{0,1,2,\ldots,m\}$ we can therefore multiply by $n+k$ and take the limit $n \to -k$ to get

$$A_k = \frac{m!}{(-k)(-k+1)\cdots(-2)(-1)(+1)(+2)\cdots(-k+m)} = \frac{m!(-1)^k}{k!(m-k)!} = (-1)^k{m\choose k}$$

Multiplying the identity above by $\frac{1}{n}$ and summing over $n=1,2,3,\ldots$ gives us the desired identity

$$\sum_{n=1}^\infty\frac{m!}{n\cdot n(n+1)(n+2)\cdots(n+m)} = \sum_{n=1}^\infty \frac{1}{n}\sum_{k=0}^m (-1)^{k}{m\choose k} \frac{1}{n+k}$$

If needed then we can further massage this into the alternative forms

$$\sum_{k=0}^m (-1)^{k}{m\choose k} \frac{H_k}{k} = \sum_{k=m+1}^\infty \frac{1}{k^2} = \zeta(2) - \sum_{k=1}^m \frac{1}{k^2} = \psi^{(1)}(m+1)$$

where $\psi^{(1)}$ is the trigamma function and $H_n$ is the harmonic number.

Winther
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  • He fixed it. Happy new year Winther. – Hazem Orabi Dec 31 '16 at 21:22
  • @HazemOrabi Thanks for pointing out the typo, I did not see it myself. Happy new year! – Winther Dec 31 '16 at 21:22
  • Briefly reading through this, what do you mean by "the limit $n \to -k$"? $k$ is a dummy variable for the summation and so this doesn't make much sense to me since $k$ isn't fixed. – Brevan Ellefsen Dec 31 '16 at 22:11
  • @BrevanEllefsen It's just abuse of notation. I edit it to avoid this mixup (thanks for the comment). It means for any fixed number $k' \in {0,1,2,\ldots,m}$ we take the limit $n\to -k'$. – Winther Dec 31 '16 at 22:13