Saudi Arabia Mathematical Competitions 2012

We use the fact that the harmonic series ${\sum }_{i=1}^{\infty }\frac{1}{i}$ diverges. This means that for any real $M$, we can find an $n$ such that ${\sum }_{i=1}^n\frac{1}{i}>M$. But ${\sum }_{i=1}^k\frac{1}{i}$ is finite for any fixed $k$, so therefore for any real $M$ and integer $k$ we can find an $n$ such that ${\sum }_{i=k+1}^n\frac{1}{i}>M$.

Take the smallest value of $m$ such that ${\sum }_{i=2012^{2012}}^m\frac{1}{i}>2012$. We claim that $n_1=2012^{2012}$, $n_2=2012^{2012}+1,\dots ,n_k=m$ is a sequence that works. We have by hypothesis that $2012<{\sum }_{i=1}^k\frac{1}{n_i}$.

Suppose that ${\sum }_{i=1}^k\frac{1}{n_i}\ge 2012+{\left(\frac{1}{2012}\right)}^{2012}$. Because $m>2012^{2012}$, we have ${\sum }_{i=1}^{k-1}\frac{1}{n_i}>2012$. But this means ${\sum }_{i=2012^{2012}}^{m-1}\frac{1}{i}>2012$, contradicting the minimality of $m$. So in fact we must have ${\sum }_{i=1}^k\frac{1}{n_i}<2012+{\left(\frac{1}{2012}\right)}^{2012}$, as desired.