Find sequences $a_n$ that converges to 1 and $b_n$ that converges to $\infty$ such that $a_n^{b_n}$ converges to 5

Question

I stuck in this question for a few hours now. Can't understand how to find one and all my guesses failed.

The Question :

Find sequence $a_n$ that converges to 1 (not constant sequence).

Find sequence $b_n$ that converges to $\infty$.

such that $a_n^{b_n}$ converges to $5$.

I found such sequences that converges to $2e$ (5.4) but it's not exactly 5..

Thank you very much and have a nice day!

Answer 1

Hint

By noting $e=\lim_{n\to\infty} (1+{1\over n})^n$, start from

$$ a_n={1+{\ln 5\over n}}\quad,\quad b_n=n $$

A simpler one:

$$ a_n=5^{1\over n}\quad,\quad b_n=n $$

Answer 2

Verify that $a_n = 1 + \dfrac{\ln(5)}{n}$ and $b_n = n$ does the trick.

Answer 3

In response to a comment by the proposer about "not using $\ln$": For $n\in \Bbb N$ we know there exists $m\in \Bbb N$ such that $(1+1/n)^m> 5$ because by the Binomial Theorem, $(1+1/n)^{5n}=1+\binom {5n}{1}(1/n)+\binom {5n}{2}(1/n^2)+...>5.$

So let $b_n$ be the $least$ $m\in \Bbb N$ such that $(1+1/n)^m>5.$ And let $a_n=1+1/n.$

By definition of $b_n$ we have $(1+1/n)^{-1+b_n}\le 5<(1+1/n)^{b_n}.$ Therefore $$0<(a_n)^{b_n}-5\le (a_n)^{b_n}-(a_n)^{-1+b_n}=(a_n)^{-1+b_n}(a_n-1)\le 5(a_n-1)=5/n.$$

So $(a_n)^{b_n}\to 5. $

We cannot have $b_{n+1}<b_n.$ Otherwise $b_{n+1}$ would be an $m\in \Bbb N$ with $m<b_n$ and $(1+1/n)^m>5,$ contrary to the definition of $b_n.$ Because $5<(1+1/(n+1))^{b_{n+1}}<(1+1/n)^{b_{n+1}}.$

So, since $b_n\le b_{n+1},$ and each $b_n\in\Bbb N,$ we can conclude that $b_n\to\infty$ by showing that there is no largest $b_n$, as follows: If $n'$ is large enough that $1+1/n'<\sqrt {1+1/n}$ then $(1+1/n')^{-2+2b_n}<(1+1/n)^{-1+b_n}\le 5.$ So we must have $b_{n'}>-2+2b_n.$ This implies $b_{n'}>b_n$ because $b_n\ge b_1>2$.

Brought to you by the Year 1500. John Napier discovered/invented logarithms in the 1500's.