I'm interested in proving the following claim:
There exists a sequence of natural numbers $\left(a_{n}\right)_{n=1}^{\infty}$ such that $$\lim_{n\to\infty}\left(1-\frac{1}{2^{n}}\right)^{a_{n}}=\frac{1}{2} $$
I've studied a fair amount of calculus and algebra, yet I've never encountered such a problem before. How should I approach this claim, or rather what tools should I read about?
Thanks!
On
Try
$$a_n=\lfloor2^n\ln2\rfloor$$
and use the inequalities $2^n\ln2-1\lt a_n\le2^n\ln2$ in the Squeeze Theorem: Since $1-{1\over2^n}\lt1$, we have
$$\left(1-{1\over2^n}\right)^{2^n\ln2}\le\left(1-{1\over2^n}\right)^{a_n}\lt\left(1-{1\over2^n}\right)^{2^n\ln2-1}$$
The left- and right-hand expressions are easily seen to tend to $(e^{-1})^{\ln2}={1\over e^{\ln2}}={1\over2}$ as $n\to\infty$.
Write $\ln 2$ as a binary number with bits $b_1, b_2, \ldots$.
As others have suggested in briefly-present answers, taking logs is the secret: you want $$ \lim_{n\to+\infty}\frac{a_n}{2^n}=\ln(2) $$
and hence $$a_n=2^n\ln(2).$$ but this choice of $a_n$ is not an integer. On the other hand, the floor of it is. So we pick
$$a_n=\lfloor 2^n\ln(2) \rfloor$$
which is just the binary "whole number" whose bits (before the decimal point, so to speak) are $b_1, b_2, \ldots, b_n$.
Then the question becomes why the limit in this case is the same as in the case without the floors.
Letting $$q_n=2^n\ln(2),$$ I'm claiming that $$ \lim \left(1-\frac{1}{2^{n}}\right)^{a_{n}}=\lim \left(1-\frac{1}{2^{n}}\right)^{q_{n}}. $$
Taking log of both sides, this is the same as $$ \lim a_n \log \left(1-\frac{1}{2^{n}}\right)= \lim q_n \log\left(1-\frac{1}{2^{n}}\right)^{q_{n}} $$ or $$ \lim (a_n-q_n) \log \left(1-\frac{1}{2^{n}}\right)= 0. $$ Now $a_n - q_n$ is between 0 and 1, and $\log \left(1-\frac{1}{2^{n}}\right) \approx -\frac{1}{2^{n}}$, so it's clear that the limit is bounded below by $$-2\frac{1}{2^{n}},$$ which goes to 0, so the squeeze lemma applies and the limit is indeed 0.