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The following serie is obviously convergent, but I can not figure out how to calculate its sum :

$$\sum \frac{6^n}{(3^{n+1}-2^{n+1})(3^n - 2^n)}$$

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    $\begingroup$ Classically the thing to do in a series like this is look for a 'telescope' - to write the general term as something of the form $f(n+1)-f(n)$. Since the denominator would arise as the difference of some expression $\dfrac{g(n+1)}{3^{n+1}-2^{n+1}}-\dfrac{g(n)}{3^n-2^n}$, you might want to work through the calculations on expanding that difference out and see whether you can find a general expression for $g()$ that makes it equal your term. $\endgroup$ – Steven Stadnicki Apr 8 '15 at 22:55
  • $\begingroup$ If I suppose that $g(n) = a^n$, then $a=6$ but this value doesn't match. What form do I have to search $g(n)$? I've tried with $g(n)=a^n -b^n$, but it s difficult to find $a$ and $b$... $\endgroup$ – Sebastien Apr 8 '15 at 23:34
  • $\begingroup$ Sebastien: keep $g(n)$ ambiguous; write out the result that you get from computing the difference I listed as a fraction of the form something divided by $(3^{n+1}-2^{n+1})(3^n-2^n)$ and see what that something comes out as in terms of $g(n)$. Setting that equal to $6^n$ then gives you an equation that you can try and solve. $\endgroup$ – Steven Stadnicki Apr 8 '15 at 23:41
  • $\begingroup$ I have easily $g(n+1)(3^n-2^n) - g(n)(3^{n+1}-2^{n+1}) = 6^n$. But I'm stuck because of the $g(n)$ and $g(n+1)$. $\endgroup$ – Sebastien Apr 8 '15 at 23:44
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    $\begingroup$ Try something like $\frac{2^{n}}{3^{n}-2^{n}}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$. $\endgroup$ – André Nicolas Apr 9 '15 at 0:04
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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

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$$\dfrac{3^n}{3^n-2^n}-\dfrac{3^{n+1}}{3^{n+1}-2^{n+1}}=\dfrac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$$ $$\dfrac{2^n}{3^n-2^n}-\dfrac{2^{n+1}}{3^{n+1}-2^{n+1}}=\dfrac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$$

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