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}})))$
?