# Proof a series is coverges to a specific sum

I need to prove that the sum of the following series is convergent to : $1 \ge Sum$

$$\sum_{n=1}^\infty \ \left[\left(\frac{2n+1}{n}\right)\left(\frac{2n+2}{n}\right)\cdots\left(\frac{2n+n}{n}\right)\right]^{-1}$$

Well I succeed to compare it to another series but could only prove its $2 \ge Sum$.

I would like to get a hint and explanation about the rationale of the solution.

Thanks!

• Each of the terms $\frac{2n+k}{n}$ is $>2$, so we have $\sum_{n=1}^\infty a_n$ where $a_n<\frac{1}{2^n}$. – almagest Apr 23 '16 at 12:04
• @OlivierOloa I think there is a tiny $]^{-1}$ at the end! – almagest Apr 23 '16 at 12:06
• @almages That's true .. watch out from the $]^{-1}$ – Barak Mi Apr 23 '16 at 12:07
• And you should know that $\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots=1$ – almagest Apr 23 '16 at 12:09
• The dummy variable is $i$ or $n$? – Olivier Oloa Apr 23 '16 at 12:11

Since $(2n+1)\ldots (2n+n)\geq (2n)^n=2^n n^n$,
$\displaystyle \sum_{n=1}^N \frac{n}{2n+1}\ldots\frac{n}{2n+n} \leq \sum_{n=1}^N \frac{1}{2^n} \leq \sum_{n=1}^\infty \frac{1} {2^n}=1$