Evaluating $\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right)$ $$\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) = \;?$$

I have been trying to see if it can be written as sum of two telescope terms but it looks tricky. Any help ?
 A: Hint:
You might compare the partial sum for $n=1$ to $k$ with $\frac11 -\frac12+\frac13 -\frac14+\cdots -\frac1{4k}$, which has a well known limit.
Added:
If you do, you will find the difference is $\left(\frac12-\frac14\right)+\left(\frac16-\frac18\right)+\cdots+\left(\frac1{4‌​k+2}-\frac1{4k}\right)$ which is half $\frac11 -\frac12+\frac13 -\frac14+\cdots -\frac1{2k}$.
So the limit of the original sum is $$\log_e 2 + \frac12 \log_e 2 \\=\frac32 \,\log_e 2.$$
A: Hint: The sum is a rearrangement of the terms of the alternating harmonic series.
Can you find a series to compare term-wise to? Note that $-\frac{1}{2n} = - \frac{1}{4n} - \frac{1}{4n}$.

Since everyone is very hectic in the comments, let me be clear here:
The sum converges. The rearrangement itself is not enough to prove this, of course, since the harmonic series converges conditionally. However, my hint that it is a rearrangement was to hint at the possibility that we can use a comparison to prove it.

Full solution
$
\sum \left( \frac{1}{4n-3} + \frac{1}{4n-1} - \frac{1}{4n} - \frac{1}{4n} \right) = \sum \left(\frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n}\right) + \sum \left( \frac{1}{4n-2} - \frac{1}{4n}\right).$
$= \ln 2 + \frac{1}{2} \sum \left( \frac{1}{2n-1} - \frac{1}{2n} \right) = \frac{3}{2} \ln 2.$
A: $$\sum\limits_{n=1}^k \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) =1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2k} + \sum\limits_{n=k}^{2k-1} \frac{1}{2n+1}$$
Can you evaluate the limits of the two series on the right-hand side?
