# 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:

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

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

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

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