# A limit

Félix Locas presented me this problem.

Let $r(n) = \lfloor \log_2 \frac{n}{\log_2 n} \rfloor$. Show that

$\lim_{n \rightarrow \infty} \left( \log 2+\sum_{k=1}^{r(n)} \frac{1}{k(k+1) 2^k} \right)^n = 1.$

## My solution

The series $\sum_{k=1}^{\infty} \frac{1}{k(k+1) 2^k}$ is easy to calculate. It is, for instance, the difference between the integrals of geometric series:

$\sum_{k=1}^\infty \frac{1}{k(k+1) 2^k} = \sum_{k=1}^\infty \frac{1}{k 2^k} - \sum_{k=1}^\infty \frac{1}{(k+1) 2^k} = 1-\log 2.$Read More »