How to check whether this sum converge or diverge? Consider the following sum:
$$\sum_{k=1}^\infty\prod_{j=1}^k\frac{1}{\sqrt{j+1}-\sqrt{j}+1}$$
How could I check whether this sum converge or diverge? Root and ratio tests are inconclusive... 
 A: You have the equivalence (for $j\gg 1$) : 
$$\sqrt{j+1}-\sqrt{j}+1 \sim 1+\sqrt{j}\left(\sqrt{1+\frac 1j}-1\right)\sim 1+\frac 1{2\sqrt{j}}-\frac 1{8j\sqrt{j}}$$
So that
$$\log\left(\sqrt{j+1}-\sqrt{j}+1\right)\sim \frac 1{2\sqrt{j}}-\frac 1{8j}$$
and
$$\prod_{j=1}^k\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\sim \exp\left[C-\sum_{j=1}^k \frac 1{2\sqrt{j}}-\frac 1{8j}\right]\sim D\;e^{-\sqrt{k}+(\ln k)/8}\sim D\;\sqrt[8]{k}\;e^{-\sqrt{k}}$$
I'll let you conclude.
A: $$
\begin{align}
\frac1{\sqrt{j+1}-\sqrt{j}+1}
&=\frac{\sqrt{j+1}+\sqrt{j}}{1+\sqrt{j+1}+\sqrt{j}}\\
&=1-\frac1{1+\sqrt{j+1}+\sqrt{j}}\\
&\le1-\frac1{3\sqrt{j+1}}
\end{align}
$$
Since $1-x\le e^{-x}$ for all $x$,
$$
\begin{align}
\prod_{j=1}^k\left(1-\frac1{3\sqrt{j+1}}\right)
&\le\exp\left(-\sum_{j=1}^k\frac1{3\sqrt{j+1}}\right)\\
&\le\exp\left(-\frac13\int_2^{k+2}\frac{\mathrm{d}x}{\sqrt{x}}\right)\\
&\le\exp\left(-\frac23\left(\sqrt{k+2}-\sqrt2\right)\right)\\
&\le e^{\frac23\sqrt2}e^{-\frac23\sqrt{k+2}}
\end{align}
$$
Noting the critical points at $u=0$ and $u=3$, we get that
$$
u^3e^{-u}\le27e^{-3}
$$
Therefore,
$$
e^{\frac23\sqrt2}e^{-\frac23\sqrt{k+2}}\le\frac{e^{\frac23\sqrt2}\,27e^{-3}}{\left(\frac23\sqrt{k+2}\right)^3}
$$
Now use the $p$-test with $p=\frac32$.
A: A different solution, which is not mine, is the following:
Let  $\alpha_{k}=\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\,,\; k\in\mathbb{N}$, then 
\begin{align*}
1-\frac{\alpha_{k+1}}{\alpha_k}&=1-\frac{\prod_{j=1}^{k+1}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}{\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}\\
&=1-\frac{1}{\sqrt{k+2}-\sqrt{k+1}+1}\\
&=\frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+2}-\sqrt{k+1}+1}\\
&=\frac{k+2-k-1}{\big(\sqrt{k+2}-\sqrt{k+1}+1\big)\big(\sqrt{k+2}+\sqrt{k+1}\big)}\\
&=\frac{1}{\sqrt{k+2}+\sqrt{k+1}+1}\,.
\end{align*}
Because 
\begin{align*}
\mathop{\lim}\limits_{k\to+\infty}\frac{k}{\sqrt{k+2}+\sqrt{k+1}+1}&=\mathop{\lim}\limits_{k\to+\infty}\frac{\sqrt{k}}{\sqrt{1+\frac{2}{k}}+\sqrt{1+\frac{1}{k}}+\frac{1}{\sqrt{k}}}=+\infty\,,
\end{align*}
by Raabe's criterion, we have that the series  $\sum_{{k}=1}^{\infty}\big(\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\big)$ converges.
