TITLE: Sum of the series $\sum u_n$ where $u_n=\frac{\sqrt{(n-1)!}}{(1+\sqrt{1}) \dots (1+\sqrt{n})}$
QUESTION [3 upvotes]: While I'm able to prove that the series $u_n=\frac{\sqrt{(n-1)!}}{(1+\sqrt{1}) \dots (1+\sqrt{n})}$ converges, I don't see the trick to compute the value of its sum starting at $n=2$.
Any clue on the way to compute the sum?

REPLY [6 votes]: Lemma: For $n>1$, one has $u_n\sqrt n -u_{n-1}\sqrt{n-1}=-u_n$.
Proof: We have
\begin{align*}
\color{blue}{u_n\sqrt n} -\color{red}{u_{n-1}\sqrt{n-1}}=\frac{\sqrt{(n-1)!}}{\prod_{k=1}^{n-1}(1+\sqrt k)}\left[\color{blue}{\frac{\sqrt n}{1+\sqrt{n}}}-\color{red}{1}\right]=-u_n.
\end{align*}
Corollary: Our sum is telescoping. Indeed,
$$\sum_{n=2}^N u_n=\sum_{n=2}^N \left(u_{n-1}\sqrt{n-1}-u_n\sqrt n\right)
=u_1-u_N\sqrt N.$$
Furthermore one can show that $u_N\sqrt{N}\to 0$ as $N\to\infty$, and therefore
$$\lim_{n\to\infty}\sum_{n=2}^N u_n=u_1=\frac12.$$
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