Help with evaluating this (likely telescopic) summation.

Question

I am trying to solve this problem which is regarding evaluating summation:

$$\sum_{k=1}^{\infty}\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}$$

Points to note:

• It seems to be telescopic summation (very likely) and I am trying to create a telescopic series by which I can directly then substitute $k$ and solve the problem.For this I currently tried to break $\frac{6^{k}}{(3^{k}-2^{k})(3^{k+1}-2^{k+1})}$ into partial fractions so dealing with them becomes simpler.
• For me second problem is dealing with $\infty$ sign as I have never before dealt with a telescopic summation where infinity is involved.So please do explain do explain how do I deal with that too.

Guys I got the solution by breaking into partial fraction.Is there some other way possible too for solving it?

Thanks a lot for help.

Answer 1

Hint: $\dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} = \dfrac{2^k}{3^k-2^k} - \dfrac{2^{k+1}}{3^{k+1} - 2^{k+1}}$

Answer 2

Hint:

$$\frac{ab}{\left(a-b\right)\left(3a-2b\right)}=\frac{3a}{2b-3a}-\frac{a}{b-a}$$

Answer 3

For your question about the $\infty$-sign:

For a sequence $(a_k)_{k \in \mathbb{N}}$ the series $\sum_{k=0}^\infty a_k$ is defined as $\lim_{n \to \infty} \sum_{k=0}^n a_k$ (if the limit exists).

So just deal with $\sum_{k=0}^n a_k$ for all $n \in \mathbb{N}$ (ie. use the hints from the other questions and the “telescope sum trick”). Then you get an expression containing some “$n$”-s. Finally compute the limit for $n \to \infty$ of that expression.