How to prove the divergence of a infinite product?

Question

I'm trying to prove the divergence of $$\prod_{k=2}^{\infty}\left(1+\frac{(-1)^{k}}{\sqrt{k}}\right)$$ which would be the same as proving the divergence of $$\sum_{k=2}^{\infty}\log\left(1+\frac{(-1)^{k}}{\sqrt{k}}\right)$$ but I'm a bit lost. Any ideas?

Cheers!

Answer 1

Join terms together in pairs: $$ \log\Bigl(1+\frac1{\sqrt{2j}}\Bigr) + \log\Bigl(1-\frac1{\sqrt{2j+1}}\Bigr) = \log\Bigl(1+\frac{\sqrt{2j+1}-\sqrt{2j}-1}{\sqrt{2j(2j+1)}}\Bigr) $$ which behaves like $-1/\sqrt{2j(2j+1)}\sim-1/(2j)$ when $j\to\infty$ (since $-1$ dominates the difference of square roots in the final numerator).

Answer 2

Near zero, $$\ln (1+X)=X-\frac {X^2}{2}(1+\epsilon (X)) $$ thus

Near infinity,

$$\ln (1+\frac {(-1)^k}{\sqrt {k}})=$$

$$\frac {(-1)^k}{\sqrt {k}}-\frac {1}{2k}(1+\epsilon (k)) $$

the first series $\sum \frac {(-1)^k}{\sqrt {k}} $ is convergent as an alternate.

the second $\sum \frac {1}{2k}(1+\epsilon (k)) $ is divergent since it is equivalent to $\sum \frac {1}{2k} $.

as a sum of a convergent and a divergent series, you series is divergent.