In mathematics, a telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation. I learnt this technique when I was doing maths olympiad. However until last year I learnt this buzz word ‘telescoping’ since I received my education in China and we call it ‘the method of differences’.
Anyway, yesterday, I was helping Po-Shen Loh to proctor this Annual Virginia Regional Mathematics Contest at CMU. Unfortunately, I did not take my breakfast. All I could digest are the seven problems. And even worse, I got stuck by the last problem which is the following.
Find \sum_{n=1}^\infty\frac{n}{(2^n+2^{-n})^2}+\frac{(-1)^nn}{(2^n-2^{-n})^2}.
It turns out that the punch line is, as you might have expected, telescoping. But it is just a bit crazy.
There is Pleasure sure, In being Mad, which none but Madmen know!
John Dryden’s “The Spanish Friar”
First consider the first term, \frac{1}{(2^1+2^{-1})^2}+\frac{-1}{(2^1-2^{-1})^2}=\frac{-2\times 2}{(2^2-2^{-2})^2}. Then take look at the second term, \frac{2}{(2^2+2^{-2})^2}+\frac{2}{(2^2-2^{-2})^2}. Aha, the sum of the first two terms becomes \frac{2}{(2^2+2^{-2})^2}+\frac{-2}{(2^2-2^{-2})^2}=\frac{-4\times 2}{(2^4-2^{-4})^2} which again will interact nicely with the 4th term (not the 3rd term). After all, the sum of the 1st, 2nd, 4th, 8th, …, n=(2^m)th terms is \frac{-4n}{(2^{2n}-2^{-2n})^2}. Note that this goes to zero.
But this only handles the terms indexed by 2’s power. Naturally, the next thing to look at is the sum of the 3rd, 6th, 12th, 24th, …, n=(3\times 2^m)th terms, which amazingly turns out have the telescoping phenomena again and is equal to \frac{-4n}{(2^{2n}-2^{-2n})^2}. Again, the this goes to zero.
We claim that starting with an odd index, say (2l+1), the partial sum of the (2l+1), 2(2l+1), \ldots, 2^m(2l+1), \ldots‘th terms goes to zero.
For people who is willing to suffer a bit more for rigorousness, the argument for the previous claim can be easily formalized by the method of differences. The key fact we have been exploiting above is the following identity.
\frac{2n}{(2^{2n}+2^{-2n})^2}+\frac{2n}{(2^{2n}-2^{-2n})^2} = \frac{4n}{(2^{2n}-2^{-2n})^2}-\frac{8n}{(2^{4n}-2^{-4n})^2}.The last bit of the ingredient is the absolute summability of the original sequence which implies that changing the order of summation does not affect the result. Hence the sum in total is 0.