$\prod\limits_{n=1}^{\infty} \left(1+\frac{(-1)^{n+1}}{n^p}\right)$ diverges when $p \leq 1/2$. It is easy to see that the infinite product $$\prod\limits_{n=1}^{\infty} \left(1+\frac{(-1)^{n+1}}{n^p}\right)$$ converges when $p>1/2$. 
My guess is that it is divergent for $p\leq 1/2$. Again it is almost obvious that it diverges for $p\leq 0$. However my question is how does one prove that this product diverges for $0<p\leq 1/2$?
 A: We have
$$\begin{eqnarray*} \prod_{n=1}^{2N}\left(1+\frac{(-1)^{n+1}}{\sqrt{n}}\right) &=& \prod_{m=1}^{N}\left(1+\frac{1}{\sqrt{2m-1}}\right)\left(1-\frac{1}{\sqrt{2m}}\right) \\&=&\prod_{m=1}^{N}\left[\left(1+\frac{1}{\sqrt{2m-1}}-\frac{1}{\sqrt{2m}}\right)-\frac{1}{\sqrt{4m^2-2m}}\right]\\&=&P_N\cdot \prod_{m=1}^{N}\left(1-\frac{1}{\sqrt{4m^2-2m}\left(1+\frac{1}{\sqrt{2m-1}}-\frac{1}{\sqrt{2m}}\right)}\right)\end{eqnarray*}$$
where 
$$ P_N = \prod_{m=1}^{N}\left(1+\frac{1}{\sqrt{2m-1}}-\frac{1}{\sqrt{2m}}\right)\sim \prod_{m=1}^{N}\left(1+\frac{C}{m^{3/2}}\right)$$
is a convergent product for $N\to +\infty$, but
$$\prod_{m=1}^{N}\left(1-\frac{1}{\sqrt{4m^2-2m}\left(1+\frac{1}{\sqrt{2m-1}}-\frac{1}{\sqrt{2m}}\right)}\right)\sim \prod_{m=1}^{N}\left(1-\frac{1}{2m-\frac{1}{2}}\right) $$
behaves like $\frac{1}{\sqrt{N}}$. The same approach can be reproduced for any $p\in\left(0,\frac{1}{2}\right)$.
A: Sketch: If $p\le 1/2,$ then for $n$ even we can pair off terms as
$$\left (1+\frac{1}{n^p}\right )\left (1-\frac{1}{(n+1)^p}\right ) = 1-\frac{1}{(n(n+1))^p} + O\left (\frac{1}{n^{1+p}}\right).$$
Note that $\sum_n O\left (\dfrac{1}{n^{1+p}}\right)$ converges. So when you apply the logarithm, the tail end of the series will be, roughly,
$$\sum_{N}^{\infty} \left (-\frac{1}{(n(n+1))^p} + O\left (\frac{1}{n^{1+p}}\right)\right) = -\infty.$$
