How to prove that $\sum_{n=1}^\infty [(\frac{2n+1}{n}) (\frac{2n+2}{n}) \cdots (\frac{2n+n}{n})]^{-1}$ converges to a sum $S \leq 1$? How to prove that $\sum_{n=1}^\infty [(\frac{2n+1}{n}) (\frac{2n+2}{n}) \cdots (\frac{2n+n}{n})]^{-1}$ converges to a sum $S \leq 1$ ?
I am trying to somehow show that this is a version of $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ that converges to $1$ but I don't see how to do that. I'm not even sure if it's the right way to go also
 A: Note that the series can be rewritten as
\begin{align*}
\sum_{n=1}^\infty\left[
\left(2+\frac{1}{n}\right)
\left(2+\frac{2}{n}\right)\cdots
\left(2+\frac{n}{n}\right)\right]^{-1}.
\end{align*}
Because $\left[
\left(2+\frac{1}{n}\right)
\left(2+\frac{2}{n}\right)\cdots
\left(2+\frac{n}{n}\right)\right]^{-1}\leq 2^{-n}$ for all $n\in\mathbb{N}$,
so
$$\sum_{n=1}^\infty\left[
\left(2+\frac{1}{n}\right)
\left(2+\frac{2}{n}\right)\cdots
\left(2+\frac{n}{n}\right)\right]^{-1}\leq
\sum_{n=1}^\infty2^{-n}=\frac{2^{-1}}{1-2^{-1}}=1.$$
Next, to show that the series converges, consider the sequence $\{S_m\}_{m=1}^\infty$ defined by
$$S_m=\sum_{n=1}^m\left[
\left(2+\frac{1}{n}\right)
\left(2+\frac{2}{n}\right)\cdots
\left(2+\frac{n}{n}\right)\right]^{-1}.$$
It is clear that $S_m$ increases, and by the preceding result that
$1$ is an upper bound of $\{S_m\}_{m=1}^\infty$. We conclude that 
$\{S_m\}_{m=1}^\infty$ converges to some number $S$, that is, 
$$\sum_{n=1}^\infty\left[
\left(2+\frac{1}{n}\right)
\left(2+\frac{2}{n}\right)\cdots
\left(2+\frac{n}{n}\right)\right]^{-1}=S.$$
A: Hint: The $n$th term looks like
$$\frac{1}{a_1\cdot a_2 \cdot \cdots \cdot a_n},$$
where each $a_k\ge 2.$
A: Notice that $2<\frac{2n+k}{n}\le 3$ for $1\le k\le n$, equality holds only if $k=n$, then
$$\frac1{3^n}<\left[\left(\frac{2n+1}{n}\right)\left(\frac{2n+2}{n}\right)\ldots\left(\frac{2n+n}{n}\right)\right]^{-1}<\frac1{2^n}$$
It follows
$$\sum_{n=1}^{\infty}\frac1{3^n}<\sum_{n=1}^{\infty}\left[\left(\frac{2n+1}{n}\right)\left(\frac{2n+2}{n}\right)\ldots\left(\frac{2n+n}{n}\right)\right]^{-1}<\sum_{n=1}^{\infty}\frac1{2^n}$$
Which means
$$\frac12<\sum_{n=1}^{\infty}\left[\left(\frac{2n+1}{n}\right)\left(\frac{2n+2}{n}\right)\ldots\left(\frac{2n+n}{n}\right)\right]^{-1}<1$$
A: We want to compute:
$$ \sum_{n\geq 1}\frac{(2n)!n^n}{(3n)!} $$
where the general term, that is positive, behaves like $\sqrt{\frac{3}{2}}\left(\frac{4e}{27}\right)^n$ due to Stirling's approximation.
Since $\frac{4e}{27}\approx\frac{2}{5}$, the series is clearly convergent.
