Spivak, Ch. 23, 6b: $f$ cont on interval around $0$, $a_n=f(1/n)$. Prove if $f'(0)$ exists, $\sum\limits_{n=1}^\infty a_n$ converges, then $f'(0)=0$ The following are the first two items from a problem in Ch. 23, "Infinite Series", from Spivak's Calculus



*Let $f$ be a continuous function on an interval around $0$, and let
$a_n=f(1/n)$ (for large enough $n$).

(a) Prove that if $\sum\limits_{n=1}^\infty a_n$ converges then
$f(0)=0$.
(b) Prove that if $f'(0)$ exists and $\sum\limits_{n=1}^\infty a_n$
converges, then $f'(0)=0$.

My question is about item (b).
The solution manual simply says

If
$$0\neq c =\lim\limits_{n\to\infty}
 \frac{a_n}{1/n}=\lim\limits_{n\to\infty} na_n$$
then by the limit comparison test, $\sum_{n=1}^\infty a_n$ wouldn't
converge, since $\sum_{n=1}^\infty 1/n$ doesn't

First of all, let me try to understand this solution.
It seems to be basically writing $a_n=(na_n)\cdot \frac{1}{n}$ and arguing that $\lim\limits_{n\to\infty} \sum_{n=1}^\infty a_n=\lim\limits_{n\to\infty}\sum_{n=1}^\infty \left [(na_n)\cdot \frac{1}{n}\right ]$ diverges if $\lim\limits_{n\to\infty} (na_n)=c\neq 0$. Is this a correct interpretation?
If so, how do we conclude that $f'(0)=0$?
 A: Considering the comment from Jochen, here is the full interpretation of what the solution manual is doing
By assumption $f'(0)$ exists. By definition, this means the following limit exists
$$f'(0)=\lim\limits_{h\to 0} \frac{f(h)-f(0)}{h}=\lim\limits_{x\to \infty} \frac{f(1/x)-f(0)}{1/x}=\lim\limits_{x\to\infty} \frac{f(1/x)}{1/x}=\lim\limits_{x\to\infty} xf(1/x)$$
Assume $\lim\limits_{x\to\infty} xf(1/x)=c\neq 0$.
Then,
$$\lim\limits_{n\to\infty} \sum_{n=1}^\infty a_n=\lim\limits_{n\to\infty} \sum_{n=1}^\infty \left [ (na_n)\cdot \frac{1}{n} \right ] = \lim\limits_{n\to\infty} \sum_{n=1}^\infty (nf(1/n))\frac{1}{n}$$
which diverges by the limit comparison test, as follows
$$\lim\limits_{n\to\infty} \frac{nf(1/n)\cdot \frac{1}{n}}{\frac{1}{n}}=\lim\limits_{n\to\infty} nf(1/n)=c\neq 0$$
Since $\lim\limits_{n\to\infty} \sum_{n=1}^\infty \frac{1}{n}$ diverges, then so does $\lim\limits_{n\to\infty} \sum_{n=1}^\infty a_n$. But this contradicts one of our initial assumptions, that $a_n$ is summable.
Hence, by proof by contradiction, we infer that
$$\lim\limits_{x\to\infty} xf(1/x)=0$$
and thus that
$$f'(0)=0$$
A: If $c=\lim_n na_n\neq0$ then for some $N$ large enough $|na_n|>|c|/2$ for all $n\geq N$. Hence
$$\frac{|c|}{2}\frac1n\leq  |a_n|,\qquad n\geq N$$
If $c\neq0$ then there is $\delta>0$ such that if $0<|h|<\delta$, then
$$c-\frac{|c|}{2}<\frac{f(h)-f(0)}{h}<\frac{|c|}{2}+c$$
There are two cases:

*

*If $f'(0)=c>0$ there is $M\geq N$ such that  $a_n<0$ (why?) for all $n\geq M$.  As $\sum_na_n$ converges, so does $\sum_n(-a_n)$ and thus $\sum_n|a_n|$ converges (at most finitely many terms $a_n$ are positive).

*If $f'(c)=c<0$, then there is $M\geq N$ such that $a_n>0$ (why?) for all $n\geq M$. Hence $\sum_n|a_n|$ converges (at most finitely many terms $a_n$ are negative).

Putting this together, yields that if $c\neq0$, then $\sum_n\frac1n$ converges which is not possible.
