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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.

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

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  • $\begingroup$ How did you break the denominator keeping the strange numerator in mind? $\endgroup$ – Devarsh Ruparelia Dec 24 '14 at 19:23
  • $\begingroup$ Use: $1\cdot 2^{k+1} = 2\cdot 2^k$ $\endgroup$ – DeepSea Dec 24 '14 at 19:25
  • $\begingroup$ I got it.Do you have some other interesting solution too except breaking into partial fractions? $\endgroup$ – Devarsh Ruparelia Dec 24 '14 at 19:26
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Hint:

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

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    $\begingroup$ enthralled by the simplification.What made you reach till here? Basically what did you think while breaking denominator keeping numerator in mind. $\endgroup$ – Devarsh Ruparelia Dec 24 '14 at 19:20
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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.

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  • $\begingroup$ I got it.I already got that part just kept it there as its the OP.Thanks for detailed explanation though. $\endgroup$ – Devarsh Ruparelia Dec 25 '14 at 8:36

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