$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $ as a rational number. $$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $$
I know from the ratio test it convergest, and I graph it on wolfram alpha and I suspect the sum is 2; however, I am having trouble with the manipulation of the fraction to show the rational number. 
ps. When it says write as a rational number it means to write the value of $S_{\infty}$ or to rewrite the fraction?
 A: Hint: $$\frac{6^k}{(3^{k+1}-2^{k+1})(3^k-2^k)}=\frac{2^k}{(3^k-2^k)}-\frac{2^{k+1}}{(3^{k+1}-2^{k+1})}$$
Now use the method of diffrences.
A: That denominator should suggest the possibility of splitting the general term into partial fractions and getting a telescoping series of the form
$$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\;,$$
where $A_k$ very likely depends on $k$. Note that if this works, the sum of the series will be
$$\frac{A_1}{3^1-2^1}=A_1\;.$$
Now
$$\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}=\frac{3^{k+1}A_k-3^kA_{k+1}-2^{k+1}A_k+2^kA_{k+1}}{(3^k-2^k)(3^{k+1}-2^{k+1})}\;,$$
so you want to choose $A_k$ and $A_{k+1}$ so that
$$3^{k+1}A_k-3^kA_{k+1}-2^{k+1}A_k+2^kA_{k+1}=6^k\;.$$
The obvious things to try are $A_k=2^k$, which makes the last two terms cancel out to leave $3^{k+1}2^k-3^k2^{k+1}=6^k(3-2)=6^k$, and $A_k=3^k$, which makes the first two terms cancel out and leaves $6^k(3-2)=6^k$; both work.
However, summing $$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\tag{1}$$ to $$\frac{A_1}{3^1-2^1}=A_1$$ is valid only if
$$\lim_{k\to\infty}\frac{A_k}{3^k-2^k}=0\;,$$
since the $n$-th partial sum of $(1)$ is 
$$A_1-\frac{A_{n+1}}{3^{n+1}-2^{n+1}}\;.$$
Checking the two possibilities, we see that 
$$\lim_{k\to\infty}\frac{2^k}{3^k-2^k}=0,\quad\text{but}\quad\lim_{k\to\infty}\frac{3^k}{3^k-2^k}=1\;,$$
so we must choose $A_k=2^k$, and the sum of the series is indeed $A_1=2$.
