Examine the uniform convergence of series

Question

I got a task: research $$\sum_{n=1}^\infty \frac{1}{n+n(n-x)^2}$$ for a uniform convergence. The clue is to check what happens when we summarize it for $$ n \in \mathbb{Z} \setminus \{0\} $$ but i have no idea how to use it.

Answer 1

If we're considering uniform convergence on $\mathbb R,$ the Weierstrass M-tesst won't work because

$$\sup_{x\in \mathbb R} |f_n(x)| = \frac{1}{n}$$

and $\sum_n 1/n = \infty.$ Nevertheless the series does converge uniformly on $\mathbb R.$

Lemma: The function $x\to \sum_{n\in \mathbb Z}\dfrac{1}{1 + (n-x)^2}$ is bounded on $\mathbb R.$

Proof: I'll leave the proof as an exercise for now, except to note that the reason for summing over all of $\mathbb Z$ is that it gives us a periodic function with period $1.$ So it's enough to show boundedness on $[0,1],$ which makes life a little easier.

On to our series: Let $\epsilon>0.$ Let $M$ be the bound on the function in the lemma. Choose $N\in \mathbb N,$ $N>M/\epsilon.$ Then for any $x,$

$$\sum_{n=N}^{\infty}\frac{1}{n+n(x-n)^2}\le \frac{1}{N}\sum_{n=N}^{\infty}\frac{1}{1+(x-n)^2} < \frac{1}{N}\sum_{n\in \mathbb Z}\frac{1}{1+(x-n)^2} < \frac{\epsilon}{M}\cdot M = \epsilon.$$

This shows the series converges uniformly on $\mathbb R.$