glycleWogry

2022-06-25

How do I evaluate this limit?
$\underset{n\to +\mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{1}{k\left(k+1\right)\cdots \left(k+m\right)}\phantom{\rule{2em}{0ex}}\left(m=1,2,3,\cdots \right)$

Aiden Norman

Rewriting the term in the sum with factorials, notice that
$\frac{1}{k\left(k+1\right)\cdots \left(k+m\right)}=\frac{1}{m!}\left(\frac{\left(k-1\right)!m!}{\left(k+m\right)!}\right).$
Then since
$\frac{\left(k-1\right)!m!}{\left(k+m\right)!}={\int }_{0}^{1}{x}^{k-1}\left(1-x{\right)}^{m}dx,$
which can be proved by induction or by using a property of the Beta Function, we see that
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\left(k+1\right)\cdots \left(k+m\right)}=\sum _{k=1}^{\mathrm{\infty }}\left(\frac{1}{m!}{\int }_{0}^{1}{x}^{k-1}\left(1-x{\right)}^{m}dx\right)$
$=\frac{1}{m!}{\int }_{0}^{1}\left(1-x{\right)}^{m}\left(\sum _{k=1}^{\mathrm{\infty }}{x}^{k-1}\right)dx=\frac{1}{m!}{\int }_{0}^{1}\left(1-x{\right)}^{m-1}dx$
$=\frac{1}{m!m}.$

dourtuntellorvl

Put ${s}_{n,m}=\sum _{k=1}^{n}\frac{1}{k\left(k+1\right)\dots \left(k+m\right)}$. We have for $m\ge 2$
$\begin{array}{rl}{s}_{n,m}& =\frac{1}{m}\sum _{k=1}^{n}\frac{k+m-k}{k\left(k+1\right)\dots \left(k+m\right)}\\ & =\frac{1}{m}\left({s}_{n,m-1}-\sum _{j=2}^{n+1}\frac{1}{j\dots \left(j-1+m\right)}\right)\\ & =\frac{1}{m}\left({s}_{n,m-1}-{s}_{n+1,m-1}+\frac{1}{1\cdots \left(1-1+m\right)}\right)\\ & =\frac{{s}_{n,m+1}-{s}_{n+1,m-1}+\frac{1}{m!}}{m},\end{array}$
and since the series $\sum _{k\ge 1}\frac{1}{k\left(k+1\right)\dots \left(k+m-1}$ is convergent, $\underset{n\to \mathrm{\infty }}{lim}{s}_{n,m+1}-{s}_{n+1,m-1}=0$ so $\underset{n\to \mathrm{\infty }}{lim}{s}_{n,m}=\frac{1}{m\cdot m!}$. This formula also works for $m=1$ by a direct computation.

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