Convergence/divergence of the series 
Prove or disprove: The series $$\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}$$ converges.

I know its general term approaches zero. But I'm unsure to move further.
Can I have a hint?
 A: A series is convergent if and only if the limit of its partial sums is finite.
Write, for $k < \infty$:
$$\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} = \sum_{n=1}^{k}\dfrac{1}{2n}+\sum_{n=1}^{k}\dfrac{(-1)^n}{n}\text{.}$$
It follows that
$$\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}  - \sum_{n=1}^{k}\dfrac{(-1)^n}{n} = \sum_{n=1}^{k}\dfrac{1}{2n}\text{.}$$

The proof: 
Suppose by contradiction that 
$$\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}$$
is finite.
We know that $$\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{(-1)^n}{n}$$
is finite, hence 
$$
\begin{align}
\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}  - \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{(-1)^n}{n} &= \lim_{k \to \infty}\left[\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} - \sum_{n=1}^{k}\dfrac{(-1)^n}{n} \right] \\
&= \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n-(-1)^n}{n} \\
&= \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{1}{2n}
\end{align}$$
will also be finite (because the sum of two finite real numbers is also finite), which is a contradiction.
A: The general term of your series is the sum of the general term of a divergent series and the general term of a convergent series. So your series diverges.
