Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie 
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?

The terms in this sum seem to be 
$$
2n\left( \dfrac {1}{2^{n-1}2n}+\dfrac {1}{3\cdot2^{n-2}2n}\right)
=
\dfrac {5}{6}\frac1{2^{n-2}}
$$
starting from $n=2$.
This is a geometric series and easy to sum.
