I don't know even how to think about it:
Let $f$ be defined in the interval $[-1, 1]$. Assume $f(0)=0$, its derivative is continuous in $(-1,1)$ and $f'(0)=\beta \neq 0$. Prove that $\sum_{n=1}^{\infty}f^2({1\over \sqrt{n}})$ diverges and $\sum_{n=1}^{\infty}f^2({1\over n})$ converges.
Hint: Use the limit comparison test (specifically, compare with the series $\sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum_{n=1}^{\infty}\frac{1}{n^2}$ .
First of all, remember that
$$\forall a\in (0,1) \exists c\in (0,a) : f(a)=f(a)-f(0)=f'(c)(a-0)=f'(c)a,$$ from where
$$\forall a\in (0,1) \exists c\in (0,a) : f^2(a)=(f'(c))^2a^2.$$
Assume wlog that $\beta>0.$ Now, since $f'$ is continuous at $0:$
$$\exists \delta>0 : |x|<\delta \implies |f'(x)-\beta|=|f'(x)-f'(0)|<\frac{\beta}{2},$$ from where
$$|x|<\delta \implies \frac{\beta}{2}=f'(x)<\frac{3\beta}{2}.$$
Thus, for $n$ big enough ($1/n<\delta$) it is
$$f^2(1/n)=(f'(c_n))^2\frac1{n^2}\le \frac{9\beta^2}{4n^{2}}$$
and ($1/\sqrt{n}<\delta$)
$$f^2(1/\sqrt{n})=(f'(c_n))^2\frac1n\ge \frac{\beta^2}{4n}.$$
From this, we can conclude what you want to show.
Since your function is continuously differentiable, then by the Taylor expansion, for $x$ sufficiently near $0$, you have $$ f(x)=\beta x+\mathcal{O}(x^2)$$ and $$ f^2(x)=\beta^2 x^2+\mathcal{O}(x^3)$$ leading to $$f^2({1\over \sqrt{n}})\sim \beta \frac1n+\mathcal{O}(\frac{1}{n^2})$$ and $$f^2({1\over n})\sim \beta^2 \frac{1}{n^2}+\mathcal{O}(\frac{1}{n^3}).$$ Consequently, $\sum f^2({1\over \sqrt{n}})$ is divergent and $\sum f^2({1\over n})$ is convergent.
