Does $\sum_{n=1}^{\infty} (-1)^n \cdot \frac{n}{2^n} $ diverge or converge? I want to prove that $\sum_{n=1}^{\infty} (-1)^n \cdot  \frac{n}{2^n}$ converges.
I was trying the ratio test:
$\sum_{n=1}^{\infty} a_n $ converges if $ q:=\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$ and diverges if $q >1$.
So we get: $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \lim_{n\to\infty} \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \lim_{n\to\infty} \frac{n2^n+2^n}{n2^{n+1}}$.
How do you calculate the limit of this function? I'm currently practicing for an exam and the solution just used this step:
$\lim_{n\to\infty} \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \frac{1}{2}(1+\frac{1}{n}) = \frac{1}{2} < 1$ therefore convergent. However I really don't see how you transform $\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}$ to $\frac{1}{2}(1+\frac{1}{n})$.

This is an alternating series, i.e., the terms alternate between positive and negative.
Also, the general term abslute value $|a_n| = \frac{n}{2^n} \to 0$ monotonically, as $n\to \infty$.
Then, by the Alternating series test, we have that the series converge.
