Does this series converge (or not) in this “simple” case?!?

Having $$\sum_{n=1}^{\infty} f \left ( {1 \over n} \right )$$ and knowing that $$f(x)$$ is always differentiable, $$f(0)=0$$ and $$f'(0)>0$$: does the series converge or not?

I was thinking about using Cauchy's criteria, due to the fact that $$f(x)$$ should be positive and tending to $$0$$ coming from the right, but not sure how to prove this...

No, it does not always converge. Try with $$f(x) = x$$.

• Oh, thanks! So, a counterexample should suffice, no need for general proof. Is this the idea? – RNani Feb 15 '19 at 13:16
• This is correct – to show that something is not always true, it suffices to provide a specific counterexample. – Minus One-Twelfth Feb 15 '19 at 13:16

With the condition $$f'(0) > 0$$ it actually always diverges:

We have (using $$f(0)=0$$):

$$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{f(h)}{h}=L > 0$$

That means for some $$\epsilon > 0$$ we have $$\frac{f(h)}{h}\ge\frac{L}2$$, when $$0 < h \le \epsilon$$. That means for $$n \ge N:=\lceil\frac1\epsilon\rceil$$ we have

$$f(\frac1n) \ge \frac{L}2\frac1n$$.

That means your series dominates a scaled harmonic series from a certain point on, so it diverges.

• Thanks. I have to digest this, but if I will have it perfectly clear I will cite it as a great answer! – RNani Feb 15 '19 at 13:34
• If something is unclear, just ask! – Ingix Feb 15 '19 at 13:34

Just set $$f(x) = x \Rightarrow$$ divergent.

In general you can conclude the divergence of the series:

You have for $$\delta$$ small enough and $$0 < x <\delta$$:

$$\frac{f(x)}{x} > f'(0)-\epsilon >0 \Rightarrow f(x) > (f'(0)-\epsilon)x$$

$$\Rightarrow \exists N \in \mathbb{N}: \sum_{n=N}^\infty f\left( \frac{1}{n}\right) > (f'(0)-\epsilon)\sum_{n=N}^\infty \frac{1}{n}$$

So, you have found a divergent minorant.