Prove $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\cdots$ converges. Consider the series:
$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\cdots$ 
Is it convergent?
I believe I need to find a way to split the terms into additive and subtraction terms, however I'm not sure how to do this and as a result prove the outcome. 
 A: The series is given by
$$1 + \sum_{k=1}^\infty (-1)^k \left(\frac1{2k} + \frac1{2k+1}\right)$$
Assuming convergence we split it up into
$$1 + \sum_{k=1}^\infty (-1)^k \frac1{2k} + \sum_{k=1}^\infty (-1)^k \frac1{2k+1}$$
Both parts converge due to leibniz and thus their sum converges as well.

Note that pulling apart series is not allowed in general, but since we were able to see that the individual parts converge, we were allowed to do it. To see the problem, look at the "identity"
$$-\ln 2 = \sum_{i=1}^\infty \frac{(-1)^i}i "=" \sum_{i=1}^\infty \frac1{2i} - \sum_{i=1}^\infty \frac1{2i-1} "=" \infty - \infty$$
A: The partial sums of 
$$1-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{6}+\frac{1}{7}\right) \cdots\tag{1}$$
converge by the Alternating Series Test.  Suppose that they converge to $a$.
The partial sum of our original series, up to the term $\pm \frac{1}{n}$, differs from a partial sum of series (1) by at most $\frac{1}{n}$. So the partial sums of the original series also converge to $a$. 
A: Hint: We have the two series
$$
\tan^{-1}(1)=1-\frac13+\frac15-\frac17+\dots
$$
and
$$
\frac12\log(2)=0+\frac12-\frac14+\frac16-\frac18+\dots
$$
which converge by the Alternating Series Test. The term-by-term difference of two convergent series is convergent.
A: \begin{align}
\dots &= \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right) - \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \dots \right) \\
&= \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right)- \frac{1}{2} \, \left(1 - \frac{1}{2} + \frac{1}{3} - \dots \right) \\
&=   \tan^{-1}(1) - \frac{1}{2} \, \log\left(1+1\right)= \frac{\pi}{4} - \frac{1}{2} \, \log(2)
\end{align}
A: Take alternating harmonic series and add $ (-2/3 ) + ( 2/4) + (-2/7) + (2/8)+......$ both alternating harmonic series and the added series are converging.
