I want to find the domain on which $f(x) = \sum_{n = 1}^\infty\frac{1}{1 + n^2x}$ is defined and converges. Now, clearly it is not defined on $S = \{\,\frac{-1}{n^2}\mid n \geq 1\,\}$, and I am wondering if it is true that $f(x)$ converges on $\mathbb R - S$. I think it is true since $\sum_{n = 1}^\infty\frac{1}{n^2}$ converges, but how can I show this? Also, does $f_N (x)= \sum_{n = 1}^N\frac{1}{1 + n^2x}$ converge uniformly as a sequence of functions?
Find the set on which $f(x)$ is defined
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real-analysis
sequences-and-series
convergence
1 Answers
1
The series converges on the set you specified excluding $\{0\}.$ This can be shown using the limit comparison test, observing that
$$\lim_{n \to \infty} \frac{1/(n^2 + 1/x)}{1/n^2} = 1$$
Convergence will not be uniform on any interval with $0$ as a boundary point since, for example with $x \in (0,\infty),$
$$\sup_{x \in (0,\infty)}\sum_{n = m+1}^{2m} \frac{1}{1 + n^2x} \geqslant \sup_{x \in (0,\infty)}\frac{m}{1 + 4m^2x} = m,$$
and the RHS diverges to infinity, violating the Cauchy criterion for uniform convergence.