# Show a series converges pointwise but not uniformly $\sum_{n=0}^\infty\frac{x^2}{n^2x+1}$

$$\sum_{n=0}^\infty\frac{x^2}{n^2x+1}$$ 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.

## 1 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 $$|S_{2N}(N) - S_N(N)| > \epsilon$$