# How do I evaluate this limit?

How do I evaluate this limit?

$$\lim_{n\to+\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)\cdots(k+m)} \qquad (m=1,2,3,\cdots)$$

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 Do you want to evaluate this for arbitrary $m$? – Chris Taylor Jan 28 '12 at 13:39 yes for arbitrary m – FrConnection Jan 28 '12 at 13:51 Note that if we multiply by $m(m!)$, we get the probability mass function of the Yule-Simon distribution with parameter $m$. (In the general Yule-Simon, $m$ is positive but not necessarily an integer, so the probability mass function is expressed in terms of the gamma function.) The value of your limit says that the Yule-Simon really is a pmf. – André Nicolas Jan 28 '12 at 16:45

Put $s_{n,m}=\sum_{k=1}^n\frac 1{k(k+1)\ldots (k+m)}$. We have for $m\geq 2$ \begin{align*} s_{n,m}&=\frac 1m\sum_{k=1}^n\frac{k+m-k}{k(k+1)\ldots (k+m)}\\ &=\frac 1m \left(s_{n,m-1}-\sum_{j=2}^{n+1}\frac 1{j\ldots (j-1+m)}\right)\\ &=\frac 1m\left(s_{n,m-1}-s_{n+1,m-1}+\frac 1{1\cdots (1-1+m)}\right)\\ &=\frac{s_{n,m+1}-s_{n+1,m-1}+\frac 1{m!}}m, \end{align*} and since the series $\sum_{k\geq 1}\frac 1{k(k+1)\ldots (k+m-1}$ is convergent, $\lim_{n\to \infty}s_{n,m+1}-s_{n+1,m-1}=0$ so $\lim_{n\to\infty}s_{n,m}=\frac 1{m\cdot m!}$. This formula also works for $m=1$ by a direct computation.
Rewriting the term in the sum with factorials, notice that $$\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m!}\left(\frac{(k-1)!m!}{(k+m)!}\right).$$ Then since $$\frac{(k-1)!m!}{(k+m)!}=\int_0^1 x^{k-1} (1-x)^m dx,$$ which can be proved by induction or by using a property of the Beta Function, we see that $$\sum_{k=1}^\infty \frac{1}{k(k+1)\cdots(k+m)}=\sum_{k=1}^\infty \left(\frac{1}{m!} \int_0^1 x^{k-1} (1-x)^m dx\right)$$ $$=\frac{1}{m!}\int_0^1 (1-x)^m\left( \sum_{k=1}^\infty x^{k-1} \right)dx=\frac{1}{m!}\int_0^1 (1-x)^{m-1}dx$$
$$=\frac{1}{m! m}.$$