, 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.