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
$$\lim_{n\to\infty}\left(1+{(-1)^n\over\sqrt n}\right)=1.$$
 A: 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 $+\infty$.
A: I would like for some approval to my answer, please.

Show that $$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$$ diverges even though$$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges.

$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges.
Indeed, by Leibnitz' converges test $$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges. Thus, $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty(\log(1+\frac{(-1)^k}{\sqrt k})+\frac{(-1)^k}{\sqrt k})$ diverges.
We observe that for all $k\geq 2$,
$$
|\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|\\
=|\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3} + \frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5}+-...|
$$
But since we know that the last series is convergen (to $|\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|$), then it is equal to
$$
=|(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-...|=:A
$$
but every addend in this series is positive because for all $l\in\mathbb{N}$:
$$
(-1)^{2l}(2l+1)\sqrt k^{2l+1}>(-1)^{2l+1}\cdot 2l\cdot\sqrt k^{2l}
$$
($\sqrt k>1$).
Thus, the previos series $A$ obtains:
$$
A=(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-...
\\ >\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}\geq \frac{1}{2k}-\frac{1}{3k^{1.5}}
$$
But because $\sum \frac{1}{2k}$ diverges and $\sum\frac{1}{3k^{1.5}}$ converges then $\sum(\frac{1}{2k}-\frac{1}{3k^{1.5}})$ diverges then $A$ diverges.
