Prove $\sum_{k=1}^\infty \frac { -2x}{(x^2+k^2)^2}$ converges uniformly using the M test.

Question

Prove $\sum_{k=1}^\infty \frac { -2x}{(x^2+k^2)^2}$ converges uniformly using the M test.

Answer 1

The bound you wrote in a comment is a good start for small $x$.

It is most convenient to split into two cases, where $|x| < 1$ and where $|x| > 1$. In the first (actually always, it's just not useful), we have

$$(x^2 + k^2)^2 \ge k^4$$ so that

$$\left|\frac{-2x}{(x^2 + k^2)^2}\right| \le \frac{2}{k^4}$$

The corresponding series converges, so we have uniform convergence on $[-1, 1]$.

For the second case,

$$(x^2 + k^2)^2 \ge 2x^2 k^2$$

by taking the middle term in the expansion only. [My goal here is to get simultaneously a high power of $k$, to force convergence, and a high power of $x$, to kill off the numerator.] Then

$$\left|\frac{-2x}{(x^2 + k^2)^2}\right| \le \frac{2|x|}{2x^2 k^2} = \frac 1 {|x| k^2} \le \frac{1}{k^2}$$

Again, the corresponding series is convergent. In fact, we could actually now just combine the two cases: We have the bound

$$\left|\frac{-2x}{(x^2 + k^2)^2}\right| \le \frac{2}{k^2}$$

for all $x$.