Show that $ \sum\limits_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}=\ln\frac43$

Question

$\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}$ 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?

Answer 1

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}

Answer 2

$$\dfrac{3^k-2^k}{k6^k}=\dfrac{(1/2)^k}k-\dfrac{(1/3)^k}k$$

Now $\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$ for $-1\le x<1$