Convergence of $\sum_{n=1}^\infty \ a_n$ when $a_n=f(\frac 1 n)$

Question

I need help with this problem: 

Let $f$ be a continous function over an interval that contains $0$. Let $a_n=f(\frac 1 n)$ (for $n$ large enough).

I've already showed that:

• If $\sum_{n=1}^\infty \ a_n$ converges, then $f(0)=0$.

• If $f'(0)$ exists and $\sum_{n=1}^\infty \ a_n$ converges, then $f'(0)=0$.

• If $f''(0)$ exists and $f(0)=f'(0)=0$, then $\sum_{n=1}^\infty \ a_n$ converges.

How solve this one:

1. Suppose that $\sum_{n=1}^\infty \ a_n$ converges. Must f'(0) exist?
2. Suppose that $f(0)=f'(0)=0$. Should $\sum_{n=1}^\infty \ a_n$ converge?

Answer 1

• For (1), consider the function given by $f(0)=0$, and $f(x) = x\sin \frac{\pi}{x}$ for $x\neq 0$.

• For (2), consider the function defined by $f(0)=0$, and $f(x) = \frac{x}{\ln x}$ for $x\neq 0$. (And recall that $\sum_{n=2}^\infty \frac{1}{n\ln n} = \infty$, e.g., by the integral test.)

Answer 2

The answer to (1) is clearly no. Let $f(x); x < 0$ be $-3x/2$ and $x>0$ be $f(x)=x^2$. Then $\sum_n f\left(\frac{1}{n}\right) = \sum_n \frac{1}{n^2}$ converges, is continuous, but $f$ is not differentiable at $x=0$.

Another example [if the above example feels like 'cheating'] would be $f(x) = x^2\left(\sin \left(\frac{1}{x^3}\right)+2\right)$.