Studying the series $ \sum \prod\limits_{k=2}^n \left(1+\frac{(-1)^k}{\sqrt{k}} \right) $ In order to study the series $\sum u_n$ where
$$u_n=\prod_{k=2}^n \left(1+\frac{(-1)^k}{\sqrt{k}} \right),
$$
I'm trying to express $$\ln u_n= \sum_{k=2}^n \ln \left(1+\frac{(-1)^k}{\sqrt{k}} \right)
$$
with asymptotic terms.
I can write 
$$
\ln \left(1+\frac{(-1)^n}{\sqrt{n}} \right)=\frac{(-1)^n}{\sqrt{n}}-\frac{1}{2n}+\frac{(-1)^n}{3n^{3/2}}+ o\left(\frac {1}{n^{3/2}} \right) 
$$
but what about the previous terms of the sum? Can I do the same for each of those terms? 
$$ n \geq N $$ 
$$ \ln(u_n)=\ln(u_N)+\ln(v_n) $$
$$ \ln(v_n)=\sum_{k=N+1}^n \ln (1+\frac{(-1)^k}{\sqrt{k}})=\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}-\frac{1}{2k}+O(\frac{1}{n^{3/2}}) $$
$$ u_n=u_N\exp(\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}-\frac{1}{2k}+O(\frac{1}{n^{3/2}})) $$
$$ u_n=u_N(1+\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}-\frac{1}{2k}+\frac{1}{2}(\sum_{k=N+1}^n \frac{1}{k}+2\sum_{N+1\leq p,q\leq n, p\neq q} \frac{(-1)^{p+q}}{\sqrt{pq}}+O(\frac{1}{n^{3/2}})))$$
$$ u_n=u_N(1+\sum_{k=N+1}^n \frac{(-1)^k}{\sqrt{k}}+\sum_{N+1\leq p,q\leq n, p\neq q} \frac{(-1)^{p+q}}{\sqrt{pq}}+O(\frac{1}{n^{3/2}})))$$
?
 A: Once you decided that you wanted to prove that the series $\sum\limits_nu_n$ diverges (you did, didn't you?), a simple approach is to try to bound $u_n$ from below. To this end, note that $\sqrt{2k+1}\geqslant\sqrt{2k}$ for every $k\geqslant1$ hence
$$
u_{2n+1}=\prod_{k=1}^n\left(1+\frac1{\sqrt{2k}}\right)\cdot\left(1-\frac1{\sqrt{2k+1}}\right)
$$
yields
$$
u_{2n+1}\geqslant\prod_{k=1}^n\left(1+\frac1{\sqrt{2k}}\right)\cdot\left(1-\frac1{\sqrt{2k}}\right)=\prod_{k=1}^n\left(1-\frac1{2k}\right),
$$
for every $n\geqslant1$. Again with the idea to use the simple lower bounds, note that $(1-\frac12x)^2\geqslant1-x$ for every $x$, hence
$$
u_{2n+1}^2\geqslant\frac14\prod_{k=2}^n\left(1-\frac1{2k}\right)^2\geqslant\frac14\prod_{k=2}^n\left(1-\frac1{k}\right)=\frac1{4n}.
$$
This proves that $u_{2n+1}\geqslant\frac12\frac1{\sqrt{n}}$. Since $u_{2n}\geqslant u_{2n+1}$, $u_{2n}+u_{2n+1}\geqslant\frac1{\sqrt{n}}$ for every $n\geqslant1$. Thus,
$$
\sum\limits_{k=2}^{2n+1}u_k\geqslant\sum\limits_{k=1}^n\frac1{\sqrt{k}}=\int\limits_1^{n+1}\frac{\mathrm dt}{\sqrt{\lfloor t\rfloor}}\geqslant\int\limits_1^{n+1}\frac{\mathrm dt}{\sqrt{t}}=2\sqrt{n+1}-2.
$$
This proves that the sequence of general term $\sum\limits_{k=2}^{n}u_k$ is unbounded hence the series $\sum\limits_nu_n$ diverges.
A: Is it normal that I cannot comment questions (I can comment my own answers) if I'm a new user?
To the point: You can say that you don't care about $k$s smaller than some $K$: You can always skip the elements of the sum with $n$ smaller than $K$ and it will not change the fact if the series converges or not. Then, if you skip factors with $k<K$ in every element of the (shortened) sum, it also won't change the convergence (it is just a non-zero constant). If $K$ is big enough ($k \ge K$) you can say that $|o(1/k^{3/2})|$ is smaller than $1/3k^{3/2}$ or something like that (moreover you can say that it is negative, but it probably will not help at all).
