glycleWogry

2022-06-25

How do I evaluate this limit?

$\underset{n\to +\mathrm{\infty}}{lim}\sum _{k=1}^{n}\frac{1}{k(k+1)\cdots (k+m)}\phantom{\rule{2em}{0ex}}(m=1,2,3,\cdots )$

$\underset{n\to +\mathrm{\infty}}{lim}\sum _{k=1}^{n}\frac{1}{k(k+1)\cdots (k+m)}\phantom{\rule{2em}{0ex}}(m=1,2,3,\cdots )$

Aiden Norman

Beginner2022-06-26Added 16 answers

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

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{k(k+1)\cdots (k+m)}=\sum _{k=1}^{\mathrm{\infty}}(\frac{1}{m!}{\int}_{0}^{1}{x}^{k-1}(1-x{)}^{m}dx)$

$=\frac{1}{m!}{\int}_{0}^{1}(1-x{)}^{m}\left(\sum _{k=1}^{\mathrm{\infty}}{x}^{k-1}\right)dx=\frac{1}{m!}{\int}_{0}^{1}(1-x{)}^{m-1}dx$

$=\frac{1}{m!m}.$

$\frac{1}{k(k+1)\cdots (k+m)}=\frac{1}{m!}\left(\frac{(k-1)!m!}{(k+m)!}\right).$

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

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{k(k+1)\cdots (k+m)}=\sum _{k=1}^{\mathrm{\infty}}(\frac{1}{m!}{\int}_{0}^{1}{x}^{k-1}(1-x{)}^{m}dx)$

$=\frac{1}{m!}{\int}_{0}^{1}(1-x{)}^{m}\left(\sum _{k=1}^{\mathrm{\infty}}{x}^{k-1}\right)dx=\frac{1}{m!}{\int}_{0}^{1}(1-x{)}^{m-1}dx$

$=\frac{1}{m!m}.$

dourtuntellorvl

Beginner2022-06-27Added 7 answers

Put ${s}_{n,m}=\sum _{k=1}^{n}\frac{1}{k(k+1)\dots (k+m)}$. We have for $m\ge 2$

$\begin{array}{rl}{s}_{n,m}& =\frac{1}{m}\sum _{k=1}^{n}\frac{k+m-k}{k(k+1)\dots (k+m)}\\ & =\frac{1}{m}({s}_{n,m-1}-\sum _{j=2}^{n+1}\frac{1}{j\dots (j-1+m)})\\ & =\frac{1}{m}({s}_{n,m-1}-{s}_{n+1,m-1}+\frac{1}{1\cdots (1-1+m)})\\ & =\frac{{s}_{n,m+1}-{s}_{n+1,m-1}+\frac{1}{m!}}{m},\end{array}$

and since the series $\sum _{k\ge 1}\frac{1}{k(k+1)\dots (k+m-1}$ 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.

$\begin{array}{rl}{s}_{n,m}& =\frac{1}{m}\sum _{k=1}^{n}\frac{k+m-k}{k(k+1)\dots (k+m)}\\ & =\frac{1}{m}({s}_{n,m-1}-\sum _{j=2}^{n+1}\frac{1}{j\dots (j-1+m)})\\ & =\frac{1}{m}({s}_{n,m-1}-{s}_{n+1,m-1}+\frac{1}{1\cdots (1-1+m)})\\ & =\frac{{s}_{n,m+1}-{s}_{n+1,m-1}+\frac{1}{m!}}{m},\end{array}$

and since the series $\sum _{k\ge 1}\frac{1}{k(k+1)\dots (k+m-1}$ 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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