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 series is convergent if and only if the limit of its partial sums is finite.

Write, for $k &lt; \infty$:
$$\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} = \sum_{n=1}^{k}\dfrac{1}{2n}+\sum_{n=1}^{k}\dfrac{(-1)^n}{n}\text{.}$$
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 
$$
\begin{align}
\lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n}  - \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{(-1)^n}{n} &amp;= \lim_{k \to \infty}\left[\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} - \sum_{n=1}^{k}\dfrac{(-1)^n}{n} \right] \\
&amp;= \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n-(-1)^n}{n} \\
&amp;= \lim_{k \to \infty}\sum_{n=1}^{k}\dfrac{1}{2n}
\end{align}$$
will also be finite (because the sum of two finite real numbers is also finite), which is a contradiction.
Suppose assume the given series converges. Then

$$\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}=\sum_{n=1}^\infty  \frac{1}{2n}+\sum_{n=1}^\infty \frac{(-1)^n}{n}=\frac{1}{2} \sum_{n=1}^\infty  \frac{1}{n}+\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
We already know $\sum_{n=1}^\infty  \frac{1}{n}$ diverges and $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges 

Also the difference of convergent series is  convergent. so $$\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}-\sum_{n=1}^\infty \frac{(-1)^n}{n}=\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n}$$ converges, which is a contradiction

Hence  $\sum_{n=1}^\infty \frac{\frac{1}{2}+(-1)^n}{n}$  diverges