# Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$

How to prove the following equality? $$\sum_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}.$$

$$\sum_{n=1}^{\infty}\frac{1}{\prod_{k=1}^{m}(n+k)}= \sum_{n=1}^{\infty}\frac{n!}{(n+m)!} = \frac{1}{m!}\sum_{n=1}^{\infty} \frac{1}{\binom{m+n}{n}}$$

Now

$$\frac{1}{\binom{m+n}{n}} = m \int_0^1 dx \, x^{m-1} (1-x)^n$$

so that the sum in question is equal to, upon reversing sum and integral

\begin{align}m \int_0^1 dx \, x^{m-1} \sum_{n=1}^{\infty} (1-x)^n &= m \int_0^1 dx \, x^{m-1} \frac{1-x}{x} \\ &= m \int_0^1 dx \, \left ( x^{m-2}-x^{m-1}\right) \\ &= m \left (\frac{1}{m-1}-\frac{1}{m}\right ) \\ &= \frac{1}{m-1} \end{align}

Putting this all together,

$$\sum_{n=1}^{\infty}\frac{1}{\prod_{k=1}^{m}(n+k)}= \frac{1}{(m-1) m!}$$

as was to be shown.

• @ Ron Gordon. Thanks very much. – user91500 Aug 27 '13 at 10:30
• @Artin: you're welcome. – Ron Gordon Aug 27 '13 at 10:44

Note that $\prod\limits_{k=1}^{m}(n+k) = \frac{(n+m)!}{n!}$, then prove by induction the explicit formula for the $s$'th partial sum $$\sum_{n=1}^{s}\frac{1}{\prod_{k=1}^{m}(n+k)} = \sum_{n=1}^{s}\frac{n!}{(n+m)!} = \frac{1}{m-1} \left( \frac{1}{m!} - \frac{(s+1)!}{(s+m)!}\right)$$ and take the limit $s \rightarrow \infty.$

Addendum. The partial fraction identity at the top can be shown by induction on $m$. The case $m=1$ is trivial. The induction step gives $$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1} \frac{1}{n+m+1} \\ = \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{m-k} \left( \frac{1}{n+k+1} - \frac{1}{n+m+1} \right).$$ The first part of the sum is $$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{m-k} \frac{1}{n+k+1} = \frac{1}{m!} \sum_{k=0}^{m-1} (-1)^k \binom{m}{k} \frac{1}{n+k+1}.$$ The second part is $$- \frac{1}{n+m+1} \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{m-k} = - \frac{1}{n+m+1} \frac{1}{m!} \sum_{k=0}^{m-1} (-1)^k \binom{m}{k} \\ = - \frac{1}{n+m+1} \frac{1}{m!} (0 - (-1)^m) = \frac{1}{n+m+1} (-1)^m \frac{1}{m!}.$$ Putting these two together completes the induction.

This partial fraction identity was also used here at MSE.

Another approach without using hidden beta and gamma functions where one gets to excercise manipulation of binomial coefficients and generating functions is to use $$\prod_{k=1}^m \frac{1}{n+k} = \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1}.$$ Now compute how often the fraction $1/q$ where $q\ge m+1$ occurs on the right side and with what coefficients. It occurrs for the first time in the term for $n=q-m$ and for the last time in $n=q-1,$ so its coefficient is (keeping in mind that $k = q-n-1$) $$\sum_{n=q-m}^{q-1} (-1)^{q-n-1} \binom{m-1}{q-n-1}.$$ Put $p=q-n-1$ in the above sum to get $$\sum_{p=q-(q-m)-1}^{q-(q-1)-1} (-1)^p \binom{m-1}{p} = \sum_{p=m-1}^0 (-1)^p \binom{m-1}{p} = \sum_{p=0}^{m-1} (-1)^p \binom{m-1}{p} = 0,$$ so for $q\ge m+1$ the sum telescopes and the contribution of the value $1/q$ is zero.

The leftover contribution is $$\sum_{q=2}^m \frac{1}{q}\sum_{n=1}^{q-1} (-1)^{q-n-1} \binom{m-1}{q-n-1} = \sum_{q=2}^m \frac{1}{q}\sum_{n=0}^{q-2} (-1)^n \binom{m-1}{n}$$

To conclude we ask where $\binom{m-1}{n}$ appears and for what values of $q$, getting $$\sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} \sum_{q=2+n}^m \frac{1}{q} = \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} \left(H_m - H_{n+1}\right).$$ Split the sum in two to obtain first, $$H_m \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} = H_m (0 - (-1)^{m-1} ) = (-1)^m H_m$$ and second, $$\sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} H_{n+1} = \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1} - (-1)^{m-1} H_m = \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1} + (-1)^m H_m$$ so the difference between the two is $$- \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1}.$$ This is the binomial transform of $H_{n+1}$ and since $$\sum_{n\ge 1} H_{n+1} z^n = f(z) = \frac{1}{z} \frac{1}{1-z} \log \frac{1}{1-z}$$ the sum has generating function $$\frac{1}{1-z} f\left(\frac{z}{z-1}\right) = \frac{1}{z} (1-z) \log \frac{1}{1-z} = \left(\frac{1}{z}-1\right) \log \frac{1}{1-z} \\= \sum_{m\ge 1} \left(\frac{1}{m+1}-\frac{1}{m}\right) z^m = -\sum_{m\ge 1} \frac{1}{m(m+1)} z^m.$$ Observe that we need the coefficient of $z^{m-1}.$ It follows that the original sum has value $$\frac{1}{(m-1)!} \frac{1}{(m-1)m} = \frac{1}{(m-1)m!}$$ as was to be shown.

I will prove the partial fraction identity in an additional answer.

• @ Marko Riedel: Thanks very much. Though i can't understand your proof but i like convergent series and calculation the number of convergence. Can you introduce a good book in this field that contains proofs? – user91500 Aug 28 '13 at 6:43
• I don't have it on my desktop but if memory serves the classic for these types of problems is "Concrete Mathematics" by D. Knuth, an absolute must read for all those interested in discrete mathematics, generating functions and special sequences. – Marko Riedel Aug 28 '13 at 18:19
• @ Marco Ridel: Thanks. – user91500 Aug 28 '13 at 18:28