Determine the limit to which $\prod_{n=2}^{\infty}\left (1+\frac{(-1)^n}{n}\right )$ converges Background: this is Arfken et al mathematical methods 12.5.4 and the answer is 1.
Using the infinite sin product we need the alternating terms in red to cancel when $\pi$ is plugged into z but I don't know how to do that:
$$\frac{\sin(z)}{z(1-z^2/\pi^2)}=\prod_{\color{red}{n=2}}^{\infty}(1-\frac{z^2}{n^2\pi^2})\overset{z=\pi}{=}\color{red}{(1-\frac{1}{2})}(1+\frac{1}{2})\color{red}{(1+\frac{1}{3})}(1-\frac{1}{3})\color{red}{(1-\frac{1}{4})}(1+\frac{1}{4})\dots$$
Alternate answers using other values for z like $\pi i/2$ or using other infinite series like cos are welcome. 

Here's a different approach notice that for $n=2k$
we have that the general term is equal to
$$\frac{2k+(-1)^{2k}}{2k}=\frac{2k+1}{2k}$$
And for $n=2k+1$ we have $$\frac{2k+1+(-1)^{2k+1}}{2k+1}=\frac{2k}{2k+1}$$
And the product of those two is exactly $1$,hence writing the partial product
$$P_{2k+2}=\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{5}{4}\cdot\frac{4}{5}\cdots\frac{2k+1}{2k}\cdot\frac{2k}{2k+1}\cdot\frac{2k+3}{2k+2}=\frac{2k+3}{2k+2}$$
and $$P_{2k+1}=1$$
So we have that $P_n\to1$ as $n\to\infty$ so the product converges to $1$.
