Convergence/divergence of the series

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 $$\sum_{n=1}^{k}\dfrac{\frac{1}{2}+(-1)^n}{n} - \sum_{n=1}^{k}\dfrac{(-1)^n}{n} = \sum_{n=1}^{k}\dfrac{1}{2n}\text{.}$$

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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.