How can we show that the function $\sum\limits_{m\in\mathbb{Z}}\frac{1}{1+(x+2\pi m)^2}$ is bounded?

How can we show that the function $$\sum\limits_{m\in\mathbb{Z}}\frac{1}{1+(x+2\pi m)^2}$$ is bounded from above as a function of $$x$$?

• Hint: Note that the function is periodic. – Tom Aug 14 at 12:42

First you have to convince yourself that, as a function in $$x$$, the sum is $$2\pi$$-periodic. i.e. the function

$$f(x) := \sum\limits_{m\in\mathbb{Z}}\frac{1}{1+(x+2\pi m)^2}.$$

This means that you only need to consider the values the sum takes for $$x \in [0,2\pi]$$, since $$f(\mathbb R) = f([0, 2\pi])$$ from the periodicity.

With this constraint on $$x$$, you can get away with assuming the "worst case scenario": that each of the denominators of the summands is as small as possible; and then compare to a simple convergent sequence that you should know!

• or just appeal to continuity – mathworker21 Aug 14 at 13:13