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.

2 thoughts on “A limit

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