<s>[INST] Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$. Looking looking for a verification of my proof that the above product diverges.
$$\begin{align}
\prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt n}\right) & =\prod_{n=1}^\infty\left(1+\frac1{\sqrt {2n}}\right)\left(1-\frac1{\sqrt{2n+1}}\right)\\
& =\prod_{n=1}^\infty\left(1-\frac1{\sqrt{n_1}}+\frac1{\sqrt {2n}}-\frac1{\sqrt{2n(2n+1)}}\right)\\
& =\prod_{n=1}{\sqrt{2n(2n+1)}-\sqrt{2n}+\sqrt{2n+1}-1\over\sqrt{2n(2n+1)}}\\
& \ge\prod_{n=1}^\infty{\sqrt{2n(2n+1)+2n+1}-\sqrt{2n}-1\over\sqrt{2n(2n+1)}},\quad\sqrt x+\sqrt y\ge\sqrt{x+y}\\
& = \prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\
& = \prod_{n=1}^\infty{2n-1\over\sqrt{2n+1}}
\end{align}$$
This last product diverges since 
$$\lim_{n\to\infty}{2n-1\over\sqrt{2n+1}}=\infty.$$
I'm suspicious because
$$\lim_{n\to\infty}\left(1+{(-1)^n\over\sqrt n}\right)=1.$$
 [/INST] Your calculation is wrong, but the conclusion is correct.  
$$ \left( 1 + \dfrac{1}{\sqrt{2n}}\right) \left( 1 - \dfrac{1}{\sqrt{2n+1}}\right) = 1 - \dfrac{1}{2n} + O(n^{-3/2})$$ 
The infinite product diverges, but to $0$, not $+\infty$.
</s>