calculate the serie $\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$ The following serie is obviously convergent, but I can not figure out how to calculate its sum :
$$\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$$
 A: Allright. Time to get some excel in here. After playing with it, it shows that the limit goes to 4. Now let's try to prove that.
Let's set up partial fractions, as first suggested by Steven.
I did: $\frac{A}{3^{n+1}-2^{n+1}}+\frac{B}{3^n-2^n}$ Adding these terms together by making a common denominator, gives the following numerator: $A3^n-A2^n+B3^{n+1}-B2^{n+1}$ or $A3^n-A2^n+3B3^n-2B2^n$ or $(A+3B)3^n+(-A-2B)2^n$ and this should be equal to the given numerator $6^n$ This gives us the following system of equations: $A+3B=2^n$ and $-A-2B=3^n$ because $2^n$ times $ 3^n$ is $6^n$ Solving for A and B gives $-2*2^n-3*3^n$ and $2^n+3^n$ respectively. Plugging in values for $n=1,2,3,4,5,....$ shows indeed a telescoping sum where the first term in the second column (the "B" column)  survives, which is $\frac{2^1+3^1}{3^1-2^1}=5$ When you would hypothetically stop, there is however another surviving term and that is the "last term" in the first (The "A") column. This term is
$\frac{-2^{n+1}-3^{n+1}}{3^{n+1}-2^{n+1}}$
If you divide every term by $3^{n+1}$ and let $n$ go to infinity, this terms results in $-1$ Therefore the Sum is 4 . I hope this sincerely helps you, Sebastien
A: You can use one of the following identities to telescope the series: $$\dfrac{3^{n+1}}{3^{n+1}-2^{n+1}}-\dfrac{3^n}{3^n-2^n}=\dfrac{2^n}{3^n-2^n}-\dfrac{2^{n+1}}{3^{n+1}-2^{n+1}}=\dfrac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$$
