<s>[INST] 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... 
 [/INST] 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.
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