Find the sum $\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$ I need to find the sum,
$$\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$$
I have tried to break the terms into partial fractions (method of differences) but am not able to do so. How to proceed?
 A: First we can try to split things into two pieces:
$$\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)} = \dfrac{A}{3^k-2^k} + \dfrac{B}{3^{k+1}-2^{k+1}}$$
So we have $A \cdot (3^{k+1}-2^{k+1}) + B \cdot (3^k-2^k) = 6^k$ which can be arranged to $3^k (3 A + B) - 2^k (2 A + B) = 6^k$. 
If we make $2A+B=0$ and $3A+B=2^k$, this equality will hold. This suggests that $B=-2A$ and we also see that $3A-2A=A=2^k$. Therefore $B=-2^{k+1}$.
$$S = \sum_{k=1}^{\infty} \left(\dfrac{2^k}{3^k - 2^k} - \dfrac{2^{k + 1}}{3^{k + 1} - 2^{k + 1}}\right)$$
Now it's looking like a telescoping series. This is more easily seen by displaying a few terms:
$$S = \left(\dfrac{2}{1} - \dfrac{4}{5}\right) + \left(\dfrac{4}{5} - \dfrac{8}{19}\right) + \left(\dfrac{8}{19} - \dfrac{16}{65}\right) + ...$$
Most of these fractions cancel each other out.
$$S = \lim_{K \rightarrow \infty} \sum_{k=1}^{K} \left(\dfrac{2^k}{3^k - 2^k} - \dfrac{2^{k + 1}}{3^{k + 1} - 2^{k + 1}}\right)= \lim_{K \rightarrow \infty} \left(2 - \dfrac{2^{K + 1}}{3^{K + 1} - 2^{K + 1}}\right) = 2$$
