Determining if a function is differentiable

Question

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

This above is a function which (with help) I have proved to be continuous on $[0, \infty)$

I now want to prove that $f$ is differentiable however I am struggling to see how.

I know that since $f$ is continuous, if is integrable which also applies to $f'$?

Furthermore the solution to this question posted (Uniform convergence of $f'$ on an interval implies locally uniform convergence of $f$) is clearly a related answer as I know that proving differentiability involves the uniform convergence of a sequence of functions.

Can anyone explain to me how the example I linked might be applied to my particular problem?

Many thanks!

Answer 1

we will be using following theorem

Let $f_n → f$ pointwise on the closed interval $[a, b]$, and assume that each $f_n$ is differentiable. If $(f'_n)$ converges uniformly on [a, b] to a function $g$, then the function f is differentiable and $f' = g.$

First note that $f_{n}(x)$ is differentiable $$f_n'(x) = \frac{-2}{(n+x)^3}$$

now we have to show that $f_{n}'(x)$ uniformly converges.we will use following theorem

For each $n \in N$, let $f_n$ be a function defined on a set $A\subset R$, and let $M_n > 0$ be a real number satisfying $|f_n (x)| ≤ M_n$ for all $x \in A$. If $\sum_{n=1}^{\infty} M_n$ converges, then $\sum_{n=1}^{\infty} f_n$ converges uniformly on A.

since $$|f_{n}'(x)| \leq \frac{2}{n^3} , \sum_{i=1}^{\infty} \frac{2}{n^3} <\infty$$

we have and uniform converges and hence $f$ is differentiable on every closed interval $[a,b] , a\geq0, b \geq 0$ and hence differentiable on $[0, \infty)$