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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 $0<p\leq 1/2$?

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    $\begingroup$ The product always goes to $0$ for any $p>0$, so in a sense, it diverges, always. $\endgroup$ – Simply Beautiful Art May 1 '17 at 16:54
  • $\begingroup$ Are you sure you have to deal with a product of logarithms and not with a series? $\endgroup$ – Jack D'Aurizio May 1 '17 at 18:03
  • $\begingroup$ @JackD'Aurizio Yes, it is a product. However there shouldn't be $\log$. My fault. Already edited. $\endgroup$ – 9Rx May 1 '17 at 18:28
  • $\begingroup$ @SimplyBeautifulArt No, the product converges to a nonzero value if $p>1/2$ That's what the OP is saying I think. $\endgroup$ – zhw. May 1 '17 at 19:03
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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 $p\in\left(0,\frac{1}{2}\right)$.

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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,

$$\sum_{N}^{\infty} \left (-\frac{1}{(n(n+1))^p} + O\left (\frac{1}{n^{1+p}}\right)\right) = -\infty.$$

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