Does the series $1-\frac12+\frac12-\frac1{2^2}+\frac13-\frac1{2^3}+\frac14-\frac1{2^4}+\frac15-\frac1{2^5}+\cdots$ converge or diverge?

Since the terms go to $0$, the series can be rewritten as $$ \lim_{n\to\infty}\sum_{k=1}^n\left(\frac1k-\frac1{2^k}\right)\tag{1} $$ Since $$ \begin{align} \sum_{k=1}^n\left(\frac1k-\frac1{2^k}\right) &=\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac1{2^k}\\[3pt] &\ge\log(n+1)-1\tag{2} \end{align} $$ the limit in $(1)$ does not exist.

Observe that $$S_{n}=\displaystyle\sum_{i=1}^\left\lceil \frac{n}{2}\right\rceil\left(\dfrac{1}{i}\right)-\displaystyle\sum_{i=1}^\left\lfloor \frac{n}{2}\right\rfloor\left(\dfrac{1}{2^i}\right)\ge\left(\displaystyle\sum_{i=1}^\left\lceil\frac{n}{2}\right\rceil\dfrac{1}{i}\right)-1$$ Since the sequence of partial sum diverges, the series also diverges.

When you look at the terms with the positive signs, it is the Harmonic Series, so that's divergent. Looking at the negative terms, it is essentially a geometric series, factor $1/2$ so that's (absolutely) convergent. Putting them together results in a divergent series. EDIT: I made the assumption that terms can be rearranged without affecting the sum of the series. One should take caution with this as this CAN change the outcome of the series' sum. In particular when the series of both positive and negative terms are represented by conditionally convergent series. However, in this case, we have a divergent series represented by the positive terms, against a absolute convergent series, represented by the negative terms. In such a situation (after little research) the result will always be a divergent series.