Pointwise and uniform convergence of the series $\sum\limits_{n=1}^{\infty} \frac{1}{1+n^2x^2},\ x>0$

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I am given the exercise: Check the pointwise and uniform convergence of the series $\sum\limits_{n=1}^{\infty} \frac{1}{1+n^2x^2},\ x>0$.

It is like that: $$n^2x^2 \leq 1+n^2x^2 \Rightarrow \frac{1}{n^2x^2} \geq \frac{1}{1+n^2x^2} $$

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

Do we conclude from this that $\frac{1}{1+n^2x^2}$ converges pointwise to a function $s(x)$ ?

Also,how can I check if the series converges pointwise to a function?

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Yes you're correct and by this way you proved the pointwise convergence of the series on $(0,+\infty)$. Now since $$\lim_{x\to 0}\sum_{n=1}^\infty\frac{1}{1+n^2x^2}$$ doesn't exist then the convergence isn't uniform on the interval $(0,\infty)$. Moreover, let $a>0$ then $$\frac{1}{1+n^2x^2}\le\frac1{a^2n^2},\quad\forall x\ge a$$ and the series $\sum_{n\ge1}\frac1{n^2}$ is convergent then the given series is uniformly convergent on every interval $[a,+\infty)$ for all $a>0$.