Determining whether a function is continuous

Question

If $$f(x)= \sum^{\infty}_{n=1}\frac{1}{(n+x)^2}$$ for $x \in [0, \infty)$

I want to determine whether or not $f$ is continuous on $[0, \infty)$

My attempt is: If we want to show that for each $c \in [0, \infty), \ f(x) \rightarrow f(c)$.

First we must prove that $f(c)$ exists within the domain of $f$, thus:

$$f(c) = \sum^{\infty}_{n=1}\frac{1}{(n+c)^2}$$ which exists $\forall c \in \mathbb{R} / {c=-n}$.

Next we must show that the limit of $f(x)$ as $x \rightarrow c$ exists and that it is equal to $f(c)$. \begin{align} \lim_{x \rightarrow c} f(x) & = \lim_{x \rightarrow c} \sum^\infty_{n=1} \frac{1}{(n+x)^2} \\ & = \sum^\infty_{n=1} \frac{1}{(n+c)^2} = f(c)\ \end{align}

Therefore we could conclude $f(x)$ is continuous in its domain.

I believe that I might have got this wrong, can anyone help? Thanks.

Answer 1

Each function $f_n:[0,\infty)\to \mathbb R $ defined by $f_n(x)=\frac{1}{(n+x)^2}$ are continuous. Then each finite sum $S_n(x)=\sum_{n=1}^{m}f_n(x)$ are also continuous on $ [0,\infty)$ $$\frac{1}{(n+x)^2}\le \frac{1}{n^2}\text{ , }\forall n\ge 1 \text{ and }x \in [0,\infty)$$

Then by Weierstrass M-test we have the series converges uniformly.

So, $S_n$ converges uniformly to $f$.

Uniform limit of continuous functions is continuous.

Hence $f(x)=\sum_{n=1}^{\infty}f_n(x) $ continuous.

Answer 2

For each $x\in[0,\infty)$, $x+n\geqslant n$, and therefore $\frac1{(x+n)^2}\leqslant\frac1{n^2}$. Therefore, by the Weierstrass $M$ test, the series $\sum_{n=1}^\infty\frac1{(x+n)^2}$ converge uniformly. Since each function $x\mapsto\frac1{(x+n)^2}$ is continuous and the convergence is uniform, the sum is continuous too.