Let . For any , we have (by monotonicity) . Then,
Since diverges, the double series diverges.
In order to prove that converges, you can use the similar upper bound .
There are other ways to prove this divergence. For instance (these are very similar to each other):
- for every , and (using a comparison with an integral) we have the well-worth-remembering estimate . Since diverges, we conclude.
- for every , , and, for :
where the before-last equality is a "Fubini" rearrangement of the finite sum (interchanging sums). Then , the first term is a divergent series, and the second term is positive, hence the divergence.
- you may know that since the double series has positive terms, re-arranging the terms does not alter the convergence/divergence of the sum. Then we can group the terms where . There are of them ( ), so that .
You can prove a theorem of comparison with a double-integral. Suppose x,y)\mapsto f(x,y)" alt="fx,y)\mapsto f(x,y)" /> is decresasing in both variables, and positive. Then for any , monotonicity gives:
Summing on , we get:
Then if the left-hand side integral diverges, so does the double-sum, and if the right-hand side integral converges, so does the double-sum.