Find a sequence satisfying a specific limit

Question

Find a positive sequence $\{a_n\}$, such that $$\lim_{n\to\infty}n\left({1+a_{n+1}\over a_n}-1\right)=1.$$

This problem arises when proving that

Given a positive sequence $\{a_n\}$, the RHS 1 in $$\varlimsup_{n\to\infty}n\left({1+a_{n+1}\over a_n}-1\right)\geq1$$ cannot be replaced by any greater number.

So, I think that we can construct such a sequence such that this limit (and thus upper limit) is exactly 1, to prove that we cannot reach a stronger result.

The solution directly gives the sequence $a_n=n\ln n$ without any rough work, and I can verify the alleged limit is indeed 1. But I wonder how to come up with such an example?

Answer 1

Let assume for instance the expression is identically equal to $1$, we would have:

$\begin{align}\dfrac{1+a_{n+1}}{a_n}-1=\dfrac 1n &\iff 1+a_{n+1}=(1+\tfrac 1n)a_n=\tfrac{n+1}{n}a_n\\ &\iff\frac 1{n+1}+\frac{a_{n+1}}{n+1}=\frac{a_n}n\end{align}$

So let's set $b_n=\dfrac{a_n}{n}$ to get the telescopic sequence $$b_{n+1}-b_n=-\frac 1{n+1}$$

Since the partial harmonic series $\sum\limits_{k=1}^n\frac 1k=H_n\sim\ln(n)$ we have that $b_n\sim-\ln(n)$ and $a_n\sim -n\ln(n)$.

Remark:

• The sign here is inverted $-n\ln(n)$ instead of $n\ln(n)$. This is due to the assumption in our premises where we forced the expression to be identically $1$. But with the less restrictive condition that only the limit should be $1$ then we can easily verify that $Cn\ln(n)$ is a positive solution for any constant $C>0$.