Does the following series
$$
g(x) = \sum_{n=1}^{\infty} \frac{1}{(nx)^2+1}
$$
converge uniformly
-
on $x \in (0,1)$
-
on $x\in (1, \infty)$
-
Calculate limit $\lim_{x \rightarrow \infty} g(x)$.
My attempts:
when $x \in (1, \infty)$ we have
$$
\sum_{n=1}^{\infty} \frac{1}{(nx)^2+1} \leq
\sum_{n=1}^{\infty} \frac{1}{n^2x^2} \leq
\sum_{n=1}^{\infty} \frac{1}{n^2}
$$
so series converge uniformly based on Weierstrass Theorem.
when $x \in (0, 1)$ we have
$$
\lim_{n \rightarrow \infty} \frac{1}{(nx)^2+1} = 0 = f(x) \\
f_n \rightarrow f \\
\text{We want to ensure that for every }\epsilon\sup_{x \in (0,1)} \left|\frac{1}{(nx)^2+1} – 0 \right | < \epsilon
$$
Lets find its extreme points by differentiating:
$$
f_n'(x) = \frac{-2n^2x}{((nx)^2 +1)^2}
$$
so the function is strictly decreasing and has may have an extremum when x = 0. But since $x = 0 \notin (0,1)$ we use limit to find values near zero:
$$
\lim_{\epsilon \rightarrow 0^{+}} \frac{1}{(n\epsilon)^2+1} = 1 \gt \epsilon
$$
So series converge uniformly only when $x > 1$.
Is that a correct reasoning? How can I find its limit? Does it equal the function $f$ it converges to so it's just 0?
Best Answer
I would do it as follows: if your series was uniformly convergent on $(0,1)$, then the sequence $\bigl(f_n(x)\bigr)_{n\in\mathbb N}$, where $f_n(x)=\frac1{(nx)^2+1}$, would converge uniformly to the null function. But it doesn't, since$$(\forall n\in\mathbb N):f_n\left(\frac1n\right)=\frac12.$$
On the other hand, if $N\in\mathbb N$, then$$g(N)=\sum_{n=1}^\infty\frac1{(Nn)^2+1}<\sum_{n=1}^\infty\frac1{(Nn)^2}<\sum_{n=N}^\infty\frac1{n^2}\to_{N\to\infty}0.$$This, together with the fact that $g$ is decreasing on $(1,\infty)$ (since its the sum of decreasing functions), shows that $\lim_{x\to\infty}g(x)=0.$