This question has been asked for sequences, but I couldn't find it for series.

Suppose $f(x) = \Sigma\_{n=1} ^ \infty \frac{1}{1+n^2x}.$ 

$f(x)$ converges on $(0,\infty)$ and on $(-\infty, 0)$ excluding $\frac{-1}{n^2} $, $\; n = 1,2,3 . . ..$ 

The question (Rudin 7.4), asks if it is continuous on $(0,\infty)$ and on $(-\infty, 0)$ excluding $\frac{-1}{n^2} $, $\; n = 1,2,3 . . ..$ .

The answer is yes, but I am having trouble proving it.
Here's my attempt

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We need to show that given $\epsilon > 0 $ , there is a $\delta$ s.t. $|y-x| < \delta$, implies $|f(y)-f(x)|<\epsilon$.

I tried the $3-\epsilon $ approach.

$$|f(y)-f(x)|< |f\_n(y)-f(y)| +|f\_n(y)-f\_n(x)| + |f\_n(x)-f(x)| $$

$$ = | \;\Sigma\_{n=1} ^ \infty \frac{1}{1+n^2y} \; - \; \frac{1}{1+n^2y} \;| 
+ | \;\frac{1}{1+n^2y} \; - \; \frac{1}{1+n^2x} \; | $$

$$+| \;\Sigma\_{n=1} ^ \infty \frac{1}{1+n^2y} \; - \; \Sigma\_{n=1} ^ \infty \frac{1}{1+n^2x} \;|
$$

Now, the middle term I can bound becuase of the continuity of the $f\_n$'s

but $f$ is a sum, so I don't know if that changes anything? can I say $f\_n(x)$ is $\epsilon $ -close to an $f$ given that $f$ is an infinite sum?? Or can I just treat $f$ like a function that $f\_n$ converges to uniformly?