# 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 :

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

• 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. – Steven Stadnicki Apr 8 '15 at 22:55
• 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 $b$... – Sebastien Apr 8 '15 at 23:34
• 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. – Steven Stadnicki Apr 8 '15 at 23:41
• 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 $g(n+1)$. – Sebastien Apr 8 '15 at 23:44
• Try something like $\frac{2^{n}}{3^{n}-2^{n}}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$. – André Nicolas Apr 9 '15 at 0:04

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