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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...

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

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

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

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