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? 
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{2^3}+\frac{1}{4}-\frac{1}{2^4}+\frac{1}{5}-\frac{1}{2^5}+\cdots$

I've be trying to figure out how to write this series symbolically so I can examine it's limit, but I'm having trouble.  So far the best I've come up with is:
$\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{2^n}-\frac{1}{2^{n+1}}\right)$
But obviously the above does not properly reproduce the series.  
 A: 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.
A: 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. 
A: 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.
A: Denote the series as $a_n;$ then clearly $\sum \mid a_n\ \mid$ is divergent. So, if $\sum a_n$ converges, then it is a conditionally convergent series.
By Riemann's rearrangement theorem on page 7, we can write $$\sum a_n=\sum a_n^++\sum a_n^-$$ without rearranging the terms, where $$a_n^+=max(a_n,0),\quad a^-_n=min(a_n,0).$$  
And, if $\sum a_n$ is conditionally convergent, then both $\sum a_n^+$ and $\sum a_n^-$ are divergent. But clearly $\sum a_n^-=\sum -\frac{1}{2^n}$ converges, so $\sum a_n$ is not conditionally convergent, and hence $\sum a_n$ diverges.  
Hope this helps.  
P.S. 
In fact we can write explicitly:
$$\begin{align}a_{2n-1}^+&=\frac{1}{n}\\
a_{2n}^+&=0\\
a_{2n-1}^-&=0\\
a_{2n}^-&=\frac{-1}{2^{n}}\end{align}$$
