Problem from Félix Locas:
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: \[\sum_{k=1}^\infty \frac{1}{k(k+1) 2^k} = \sum_{k=1}^\infty \frac{1}{k 2^k} - \sum_{k=1}^\infty \frac{1}{(k+1) 2^k} = 1-\log 2.\]
Furthermore, abbreviating \(r = r(n)\), \[r^{3/2} 2^{r} \sum_{k=r+1}^\infty \frac{1}{k(k+1) 2^{k}} \le \sum_{k=0}^\infty \frac{1}{\sqrt{r} 2^k} \xrightarrow{r \rightarrow \infty} 0\]
implies that for \(n\) sufficiently large we have \(\sum_{k=r+1}^\infty \frac{1}{k(k+1) 2^{k}} < r^{-3/2} 2^{-r}\) and \[\log 2+\sum_{k=1}^{r(n)} \frac{1}{k(k+1) 2^k} = 1-\sum_{k=r+1}^{\infty} \frac{1}{k(k+1) 2^k} \geq 1 - r^{-3/2} 2^{-r}. \qquad (*)\]
Finally, since \(r = \log_2 \frac{n}{\log_2 n} - \varepsilon_n\) for some \(0 \le\varepsilon_n < 1\), we have \[n r^{-3/2} 2^{-r} = \frac{2^{\varepsilon_n}\log_2 n}{(\log_2 n - \log_2 \log_2 n - \varepsilon_n)^{3/2}} \rightarrow 0\]
which implies that \[\left( 1 - r^{-3/2} 2^{-r} \right)^n \rightarrow 1.\]
Since also \(\log 2+\sum_{k=1}^{r(n)} \frac{1}{k(k+1) 2^k} \le 1\) comparing this with \((*)\) yields \[\lim_{n \rightarrow \infty} \left( \log 2+\sum_{k=1}^{r(n)} \frac{1}{k(k+1) 2^k} \right)^n = 1.\]