calculus - Infinite sum convergence test.

I have to tell if the sum $$\sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n + (-1)^{n+1}}$$
Converges or not. Can i use a comprasion test here? To assume that my sum is bigger than $\sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n + 1}$? I dont think that's OK, but that's the only way I can think of right now.

Answer

How about this? Write out the series:
$$-1(\frac{1}{1 + 1}) + 1(\frac{1}{2-1})-1(\frac{1}{3 + 1})+1(\frac{1}{4-1})-1(\frac{1}{5+1})....$$
$$=\frac{-1}{2} + 1 -\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\frac{1}{5}-\frac{1}{8}....$$

Then reorder terms...
$$1 - \frac{1}{2} + \frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}...$$
And it was the alternating harmonic series all along! Which has been proven to equal
$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}=ln(2)$$