If $\sum a_n$ converges absolutely, then so does $\sum f(a_n).$

Question

Hi I am stuck on this problem. (please dont give the solution I just need some help to formalism my solution) Let f: R to R be differentiable, with continuous derivatives and f(0) = 0 If $\sum{a_{n}}$ is an absolutely convergent series then $\sum{f(a_{n})}$ is also absolutely convergent.

I have tried using Lipschitz estimate to do this here are my thoughts/attempt, any suggestions and pieces I am missing to tie it together would be greatly appreciated

choose $k \in (a_{n})$ so that $|f'(k)|$ is an upper bound of $f(x)$ then... $$ |f(0) - f(a)| \leq |0-a|sup_{x \in (0,b)} |f'(k)| $$ Let $sup_{x \in (0,b)} |f'(x)|$ be a constant M $$ |f(a)| \leq M|a|$$ then $$\sum |f(a_{n})| \leq M \sum |a_{n}| $$

Answer 1

I like what you have so far. You know that $f$ is Lipschitz (which is the only condition you really need), so some $M$ exists such that $|f(x) - f(y)| \le M|x - y|$. As you've done, take $y = 0$ to get $$|f(x)| \le M|x|.$$ Now, note that the sequence $\sum_{k=1}^n |f(x_k)|$ is monotone increasing and bounded above by the (finite) quantity $M \sum_{k=1}^\infty |x_k|$. By the monotone convergence theorem, we get convergence of the series $\sum_k |f(x_k)|$.

Answer 2

All we need is for $f'(0)$ to exist; we don't need the existence of $f'(x)$ for any other $x.$

Proof: Since the convergence of $\sum |a_n|$ implies $a_n\to 0,$ we have

$$\left |\frac{f(a_n)-f(0)}{a_n-0} \right| = \left |\frac{f(a_n)}{a_n} \right| \le |f'(0)| + 1$$

for large $n.$ For such $n$ it follows that $|f(a_n)|\le (|f'(0)| + 1)|a_n|.$ The convergence of $\sum |f(a_n)|$ then follows from the comparison test.