Sum of the series $\sum u_n$ where $u_n=\frac{\sqrt{(n-1)!}}{(1+\sqrt{1}) \dots (1+\sqrt{n})}$

Question

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?

Answer 1

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 $$\lim_{n\to\infty}\sum_{n=2}^N u_n=u_1=\frac12.$$

Answer 2

$$\begin{eqnarray*} \log(u_n) &=& \sum_{k=1}^{n-1}\log(\sqrt{k})-\sum_{k=1}^{n}\log(1+\sqrt{k})\\&=&-\log(1+\sqrt{n})-\sum_{k=1}^{n-1}\log\left(1+\frac{1}{\sqrt{k}}\right)\\&\leq&-\log(1+\sqrt{n})-\log(2)\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}\end{eqnarray*}$$ hence $u_n$ behaves like $4^{-\sqrt{n}}$, leading to a clearly summable sequence.