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 $n$th term appears to be $$2(n+1)\left(\frac1{2^{n+1}(n+1)}+ \frac1{2^{n}3(n+1)}\right).$$
Canceling factors of $2(n+1)$, we can rewrite the $n$th term as
$$\left(\frac{1}{2^n} + \frac{1}{2^{n-1}{3}}\right) = \left(\frac{3}{2^n3} + \frac{2}{2^{n}{3}}\right) = \left(\frac{5}{2^n3}\right)$$
So the series is $$\sum_{n=1}^\infty \frac53 \cdot 2^{-n} = \frac53\sum_{n=1}^\infty 2^{-n} = \frac53.$$
