Romanian Mathematical Olympiad

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

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

(1) If $f(0)=0$, then $a_n\le 3M$, whatever $n\in {\mathbb{N}}^{\ast }$; 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 $n\in {\mathbb{N}}^{\ast }$, so the sequence $(a_n{)}_{n\ge 1}$ is unbounded.