$$\sum_{n=0}^\infty\frac{x^2}{n^2x+1}$$
On the interval $(0,\infty)$. To show pointwise convergence, I used limit comparison test with $b_n=\dfrac{1}{n^2}$. Then if $a_n=\dfrac{x^2}{n^2x+1}$, we have $\lim_\limits{n\rightarrow\infty}\dfrac{a_n}{b_n}=x$. Since $x$ is a non-zero real number, $\sum a_n$ converges iff $\sum b_n$ converges. And $\sum b_n$ converges.
To show that it is not uniformly convergent I was thinking of using the Cauchy Criterion since I don't know what this function converges to pointwise.
So here is my attempt:
Denote the $n$-th partial sum by $S_n$.
I want to show that $\exists\epsilon>0$ such that for all $N\in\mathbb{N}$ there exist $m,n>N$ and $x\in (0,\infty)$ such that $|S_n-S_m|\geq\epsilon$.
So I chose randomly $\epsilon=1$. Let $n=m+1$. Then
$$|S_n-S_m|=\left|\sum_{k=m+1}^n\frac{x^2}{k^2x+1}\right|$$
So I want to find when
$\dfrac{x^2}{k^2x+1}\geq 1$.
This is equivalent to finding $x$ such that $x^2-k^2x-1\geq 0$. And there is a solution to this since the parabolla is pointing upwards. I don't feel too comfortable since I chose $\epsilon$ arbitrarily. I would appreciate any input.
Best Answer
Good idea is always to look at the function under sum sign (in our case $\frac{x^2}{n^2x+1}$ ) and ask yourself, whether we can find such $x(n)$, for which series $\sum_{n=1}^{\infty} \frac{x^2(n)}{n^2x(n) +1}$ would diverge (it isn't formal yet, but a good point to start). In this case, we can just take $x(n) = n$, and it seems to work. So try to formalize our intuition. Denote $S_N(x) = \sum_{k=1}^{N} \frac{x^2}{n^2x+1}$.
For a moment fix $N \in \mathbb N$. Consider $|S_{2N}(N) - S_{N}(N)|$ (plugging $x=N$ is what we "find out" earlier. We cannot take different $x$'es for different terms in a sum, but try to approximate $x(n) = n$ for every $n \in \{N+1, ... ,2N\}$ by $x=N$.
We get: $|S_{2N}(N) - S_{N}(N)| = \sum_{n=N+1}^{2N} \frac{1}{n^2\frac{1}{N} + \frac{1}{N^2}} \ge \sum_{n=N+1}^{2N} \frac{1}{4N + \frac{1}{N^2}} = \frac{N}{4N + \frac{1}{N^2}} \to \frac{1}{4} $
So for example, for $\epsilon = \frac{1}{5}$, we can for any $M \in \mathbb N$, find such $N \in \mathbb N$, $N > M$, that $|S_{2N}(N) - S_N(N)| > \epsilon$