I have never worked with infinite products before, so I have no idea how to do this. So a hint would be much appreciated.
show that the product $\prod (1+\frac{(-1)^k}{\sqrt{k}})$ diverges
let $e_k = 0$ if $k$ is odd and $1$ otherwise. let $b_k = \frac{e_k}{k} + \frac{(-1)^k}{\sqrt{k}}$. show that the product $\prod (1+b_k)$ converges.
$$(1)$$
Let $\zeta_n=1-\left(1+\dfrac{1}{\sqrt{2n}}\right)\left(1-\dfrac{1}{\sqrt{2n+1}}\right)$ and we have:
$$\prod_{n= 2}^{\infty}\left(1+\frac{(-1)^n}{\sqrt{n}}\right)=\prod_{n= 1}^{\infty}(1-\zeta_n)$$
For $n\geq 1$ we have $\dfrac{1}{4n}<\zeta_n$ hence $\displaystyle\sum_{n\geq 1}\zeta_n$ diverges. Observe also that $\displaystyle\lim_{n\to\infty}\zeta_n=0$
Now since $0<\zeta_n<1\Rightarrow -\ln (1-\zeta_n)>0$ we may invoke the limit comparison test:
$$\lim_{n\to \infty}\frac{-\ln (1-\zeta_n)}{\zeta_n}=1$$
So $\displaystyle\sum_{n\geq 1}-\ln (1-\zeta_n)$ diverges and therefore $\displaystyle\sum_{n\geq 1}\ln (1-\zeta_n)$ also.
It follows that $\displaystyle\prod_{n=1}^{\infty}(1-\zeta_n)$ diverges.
$$(2)$$
Similar trick here, we now consider $\displaystyle\xi_n=1-\left(1-\dfrac{1}{\sqrt{2n+1}}\right)\left(1+\dfrac{1}{2n}+\dfrac{1}{\sqrt{2n}}\right):$
$$\prod_{n=2}^{\infty} (b_n+1)=\prod_{n=1}^{\infty}(1-\xi_n)$$
For $n\geq 1$ we have $0<\xi_n<\dfrac{1}{n^{3/2}}$ hence $\displaystyle\sum_{n\geq 1}\xi_n$ converges. Also, $\displaystyle\lim_{n\to\infty}\xi_n=0$
As before: $$\lim_{n\to \infty}\frac{-\ln (1-\xi_n)}{\xi_n}=1$$
Hence $\displaystyle\sum_{n\geq 1}\ln(1-\xi_n)$ converges and therefore $\displaystyle\prod_{n=2}^{\infty}(b_n+1)$ converges.