Convergence of a series of $f(|a_n|)$, when $f$ is continuously differentiable and $a_n$ an absolutely convergent series

Question

Question

Let $f:[0,\infty) \rightarrow \mathbb{R}$ be a continuously differentiable function, $f(0) = 0$ and $\sum_{n=1}^\infty a_n$ be an absolutely convergent series. Does $\sum_{n=1}^\infty f(|a_n|)$ converge absolutely?

What I have tried so far

I have read that a function that is continuously differentiable on $[a,b]$ is also Lipschitz continuous on $[a,b]$, but I can't confirm nor deny that claim. Maybe some of you can enlighten me. This would of course make the proof much easier!

As a hint, we were told to use the Extreme Value Theorem (EVT) which is as follows:

If $f$ is continuous on $[a,b] \rightarrow \exists c,d \ \ \ | \ \ f(c) \geq f(x) \geq f(d) \ \ \ \ \forall x \in [a,b]$.

Since $\sum_{n=1}^\infty a_n$ is convergent, that means $\lim \limits_{n \to \infty}a_n = 0$.

So as we go along our indices, $f(|a_n|)$ tends to $f(0) = 0$. Now if I knew anything about the behavior of $f$ close to $0$, say if $0$ were to be a minimum or something, I could start with $d = 0 $ and $f(d) = 0$ in the EVT, although I also don't see how that might help.

That's basically it, I haven't been able to do anything more.

Answer 1

Let $d=f'(0)$. Since $\lim_{n\to\infty}a_n=0$, you have $\lim_{n\to\infty}|a_n|=0$, and therefore$$\lim_{n\to\infty}\frac{\left|f\bigl(|a_n|\bigr)\right|}{|a_n|}=|d|.$$So, by the limit comparison test, since the series $\sum_{n=0}^\infty|a_n|$ converges, so does the series $\sum_{n=0}^\infty\left|f\bigl(|a_n|\bigr)\right|$.

Answer 2

Here another proof. Since $(a_n)$ converges to zero, a fortiori it is a bounded sequence, say $|a_n| \leq t$ for all $n.$ And since $f'$ is continuous, it is bounded by $k$ on $[0, t].$ Then, $f$ is $k$-Lispchitzian on $[0, t]$ (by MVT). Thus $|f(|a_n|)| = |f(|a_n|) - f(0)| \leq k |a_n|,$ and so the series $(f(|a_n|))$ also converges.