Prove $\prod_1^\infty (1+p_n)$ converges Let $p_{2n-1} = \frac{-1}{\sqrt{n}}$, and $p_{2n} = \frac{1}{n}+\frac{1}{\sqrt{n}}$. 
Prove $\prod_1^\infty (1+p_n)$ converges.
By numerical simulations, it appears to converge (to something around $0.759$). However, I'm not sure how to prove this. I know we can skip the first term since it's $0$. Then we can write it in the following form.
\begin{align*}
\prod_1^\infty (1+p_n) &= \prod_1^\infty \left(1+\frac{1}{2n}+\frac{1}{\sqrt{2n}}\right)\left(1-\frac{1}{\sqrt{2n+1}}\right)  \\
&=  \left(1+\frac{1}{2}+\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{3}}\right)\left(1+\frac{1}{4}+\frac{1}{\sqrt{4}}\right)\left(1-\frac{1}{\sqrt{5}}\right)...
\end{align*}
Any thoughts?

I'm not 100% sure if this can be made rigorous, but maybe you can take it as a starting point.
\begin{align*}\log \Pi(1+p_n)= \sum \log((1+\frac{1}{2n}+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n+1}}))\\\approx\sum \log((1+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n}}))\approx\sum \log(1-\frac{1}{2n})\approx \sum -\frac{1}{2n} \rightarrow \infty\end{align*}
using
$\log(1+\frac{1}{n})\approx \frac{1}{n}$

Actually there appears to be a cancellation that makes it so I can't neglect $1/2n$ the way I did.  $(1+\frac{1}{2n}+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n+1}}) =1+(\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}})+(\frac{1}{2n}-\frac{1}{\sqrt{(2n)(2n+1)}})-\frac{1}{2n\sqrt{2n+1}}$ and asymptotically the first grouped terms are $\approx \frac{1}{2(2n)^{3/2}}$ ... in short, the product converges.
