Show that $ \sum\limits_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}=\ln\frac43$ 
$\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}$ i must prove this sum converges to $\ln(4/3)$.

i tried to write expresion :$\frac{3^k-2^k}{k\cdot6^k}=\frac 1 k \left(\frac 1 {2^k-3^k}\right)$ and make two sums but i tink this sums is a dezvoltation in series of  a function. $\ln(4/3)=\sum_{k=1}^\infty \frac{(-1/3)^k} k$ => we must prove $\sum_{k=1}^\infty (-1/3)^k$ is equal with $\sum_{k=1}^\infty (1/2^k-3^k)$ and that is not inductive and that is hard to. Have any ideea?

Work with two sums separately:
\begin{align}
\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k} & = \sum_{k=1}^\infty \frac{(3/6)^k} k - \sum_{k=1}^\infty \frac{(2/6)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(1/2)^k} k - \sum_{k=1}^\infty \frac{(1/3)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(-1)^k(-1/2)^k} k - \sum_{k=1}^\infty \frac{(-1)^k(-1/3)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(-1)^k(x-1)^k} k - \sum_{k=1}^\infty \frac {(-1)^k(y-1)^k} k \text{ where } x = \frac 1 2 \text{ and } y = \frac 2 3 \\[10pt]
& = \ln x - \ln y   \\[10pt]
& = \ln \frac 1 2 - \ln \frac 2 3 = \ln \frac{1/2}{2/3} = \ln \frac 3 4
\end{align}
