Spivak: Calculation of $\lim\limits_{n\to \infty }na_n $ when $\sum\limits_{n=1}^\infty a_n$ converges

Question

This is basically a question based on Problem 6(b) of Chapter 23 in Spivak's Calculus. I have to prove that if the following are met:

• $f$ is continuous on an interval around $0$

• $a_n=f\left(\frac{1}{n}\right)$ (for n large enough)

• $\sum\limits_{n=1}^\infty a_n$ converges

• $f'(0)$ exists

  then $f'(0)=0$

I have already proven in part (a) that $f(0)=0$ and for part (b) I tried to work like this: $f'(0)= \lim\limits_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim\limits_{n\to \infty }nf(1/n)=\lim\limits_{n\to \infty }na_n $.

I tried to calculate $\lim\limits_{n\to \infty }na_n $ by claiming that $a_n$ is bounded(which I think is correct) but I seem to be missing something.

Any help?

Answer 1

HINT: You're told that $f'(0)$ exists. Suppose it is $L\ne 0$. Using what you've said, what does that tell you about $a_n$ for large $n$?