I'm given the following problem (Putnam 1984/A2):
Evaluate $$\sum_{n=1}^{\infty}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$$
The solution I've been able to find suggests that performing an 'easy induction' shows us that the $n$th partial sum for this series is $$ \frac{6(3^n-2^n)}{(3^{n+1}-2^{n+1})}$$
When we take the limit of the $n$th partial sum as $n$ becomes arbitrarily large, we find that the series converges to 2.
I have 2 questions. 1) What's meant by an 'easy induction', and, 2) How does one find the $n$th partial sum in this case?
We want to prove $$\sum_{n=1}^{k}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}=\frac{6(3^k-2^k)}{3^{k+1}-2^{k+1}}$$ We can check that for $k=1$, $$\frac{6^1}{(3^{1+1}-2^{1+1})(3^1-2^1)}=\frac{6(3^1-2^1)}{3^{1+1}-2^{1+1}}$$ which is true because $3^1-2^1=1$. Then assume it is true up to $k$ and verify it for $k+1$ $$\sum_{n=1}^{k+1}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}=\frac{6(3^k-2^k)}{3^{k+1}-2^{k+1}}+\frac{6^{k+1}}{(3^{k+2}-2^{k+2})(3^{k+1}-2^{k+1})}\\ =\frac{6(3^k-2^k)(3^{k+2}-2^{k+2})+6^{k+1}}{(3^{k+2}-2^{k+2})(3^{k+1}-2^{k+1})}\\ =\frac{6(3^{2k+2}+2^{2k+2}-6^k(3^2+2^2))+6^{k+1}}{(3^{k+2}-2^{k+2})(3^{k+1}-2^{k+1})}\\= \frac{6(3^{2k+2}+2^{2k+2}-2\cdot6^{k+1})}{(3^{k+2}-2^{k+2})(3^{k+1}-2^{k+1})}\\ =\frac{6(3^{k+1}-2^{k+1})^2}{(3^{k+2}-2^{k+2})(3^{k+1}-2^{k+1})}\\ =\frac{6(3^{k+1}-2^{k+1})}{(3^{k+2}-2^{k+2})}$$ Maybe "an easy induction" is in the eye of the beholder.