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 $x$?

  • 2
    $\begingroup$ Hint: Note that the function is periodic. $\endgroup$ – 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!

  • $\begingroup$ or just appeal to continuity $\endgroup$ – mathworker21 Aug 14 at 13:13

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