The following is a problem from Ch. 23 of Spivak's Calculus

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

6(d) Suppose $\sum\limits_{n=1}^\infty a_n$ converges. Must $f'(0)$ exist?

The answer I came up with is simply a function that has a kink at $f(0)=0$ (ie, not differentiable there), but which for non-negative $x$ is such that $a_n=f(1/n)$ and $\{a_n\}$ is summable.

For example, $f(x)=x^2$ for $x\geq 0$, and $f(x)=-x$ for $x<0$.

Then, $f(0)=0$, $f$ is continuous there, and $f'(0)$ does not exist, but $\{a_n\}=\{1/n^2\}$ is summable.

Is this an adequate counterexample to the idea that $f'(0)$ must exist if $\sum a_n$ converges?

Then I looked at the solution manual and it says

No. Consider $f(x)=x\sin\{(\pi x)\}$, $f(0)=0$.

But isn't this function differentiable at $0$?

  • $\begingroup$ Please check that my edit preserves the intent, and compare with the question title. $\endgroup$
    – abiessu
    Commented Nov 1, 2022 at 17:24
  • 2
    $\begingroup$ Please explain how the series relates to $f$, e.g. by quoting the relevant part of the exercise. $\endgroup$
    – Joe
    Commented Nov 1, 2022 at 17:27
  • $\begingroup$ I forgot to put in the beginning of the problem statement. It is fixed now. $\endgroup$
    – xoux
    Commented Nov 1, 2022 at 17:34
  • 1
    $\begingroup$ IT must be a typo in your booklet. I think the meant to write $x\sin(\pi/(x))$ for then $f(1/n)=0$ $\endgroup$
    – Mittens
    Commented Nov 1, 2022 at 18:01
  • $\begingroup$ $\frac{d}{dx}(x \sin(π x)) = \sin(π x) + π x \cos(π x)$ $\endgroup$
    – jimbo
    Commented May 31, 2023 at 3:35

1 Answer 1


I believe the question is hinting at a comparison between continuous and discrete limits.

The answer is likely a typo that intended $f(x)= x\sin(\frac{\pi}{x})$

Here $f(\frac{1}{n})=0$ for all $n$, yet $f$ is not differentiable at 0.

  • 1
    $\begingroup$ @jordi, thanks! $\endgroup$
    – Dunham
    Commented Jun 1, 2023 at 1:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .