Romanian Mathematical Olympiad

a) The continuous function $\varphi : (0, 1] \to \mathbb{R}$, $\varphi(h) = \frac{I(h) - 2f(0)h}{h^2}$, may be prolonged by continuity at $0$, since
$$\begin{align*} \lim_{h \to 0} \varphi(h) &= \lim_{h \to 0} \frac{(I(h) - 2f(0)h)'}{2h} = \lim_{h \to 0} \frac{f(h) + f(-h) - 2f(0)}{2h} = \\ &= \frac{1}{2} \lim_{h \to 0} \left( \frac{f(h) - f(0)}{h} - \frac{f(-h) - f(0)}{-h} \right) = \frac{1}{2} (f'(0) - f'(0)) = 0. \end{align*}$$
Therefore $\varphi$ is bounded on $(0, 1]$.
Let $M = \sup\{|\varphi(h)| : 0 < h \le 1\}$. Then $|I(h) - 2f(0)h| \le Mh^2$, whatever $h \in (0, 1]$; the inequality obviously also holds for $h = 0$.

b) Since
$$||I(h)| - 2|f(0)|h| \le |I(h) - 2f(0)h| \le Mh^2, \quad 0 < h \le 1,$$
it follows that
$$2|f(0)|h - Mh^2 \le |I(h)| \le 2|f(0)|h + Mh^2, \quad 0 < h \le 1,$$
whence
$$\frac{2|f(0)|}{\sqrt{k}} - \frac{M}{k\sqrt{k}} \le \sqrt{k} |I(1/k)| \le \frac{2|f(0)|}{\sqrt{k}} + \frac{M}{k\sqrt{k}}, \quad k \in \mathbb{N}^*.$$
But
$$\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \ge \sum_{k=1}^{n} \frac{1}{\sqrt{n}} = \sqrt{n}, \quad n \in \mathbb{N}^*.$$
and
$$\begin{align*} \sum_{k=1}^{n} \frac{1}{k\sqrt{k}} &= 1 + \sum_{k=2}^{n} \frac{1}{k\sqrt{k}} \le 1 + 2 \sum_{k=2}^{n} \left( \frac{1}{\sqrt{k-1}} - \frac{1}{\sqrt{k}} \right) = \\ &= 1 + 2 \left( 1 - \frac{1}{\sqrt{n}} \right) < 3, \quad n \in \mathbb{N}^*. \end{align*}$$
According with these inequalities, it follows that

(1) If $f(0) = 0$, then $a_n \le 3M$, whatever $n \in \mathbb{N}^*$; since the sequence $(a_n)_{n \ge 1}$ is increasing, it is therefore convergent.

(2) If $f(0) \ne 0$, then $a_n \ge 2|f(0)|\sqrt{n} - 3M$, whatever $n \in \mathbb{N}^*$, so the sequence $(a_n)_{n \ge 1}$ is unbounded.