What Is the Infimum of This Set of Superior Limits?

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Let $\mathbb{R}^\omega_{+}$ be the set of all sequences $(a_n)_{n=1}^\infty$ of positive real numbers. Determine the infimum $$\inf\left\{\limsup_{n \to \infty} \left(\frac{1 + a_{n+1}}{a_n}\right)^n : (a_n) \in \mathbb{R}^\omega_{+}\right\}$$

 

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Spoiler: Hint

First show, by way of contradiction, that $e$ is a lower bound of the set of superior limits. Then show that $e$ is the optimal bound.

 

[billtodd]

Let $\mathbb{R}^\omega_{+}$ be the set of all sequences $(a_n)_{n=1}^\infty$ of positive real numbers. Determine the infimum $$\inf\left\{\limsup_{n \to \infty} \left(\frac{1 + a_{n+1}}{a_n}\right)^n : (a_n) \in \mathbb{R}^\omega_{+}\right\}$$

Does a proof by definition of limsup and infimum of a set also work? I mean not through contradiction as you suggest.

For example one can look at the sequence $a_n=n/x$, and see that $((1+a_{n+1})/a_n)^n$ converges to $e^{x+1}$, where $x>0$, since we take the infinimum over all limsup, then when $x\to 0$ we get the infimum here.
This is seen as the optimum.
But I don't see how show that every other positive sequence gives a bigger limsup.

 

[Office_Shredder]

I have some thoughts on progressing towards e being the infimum. Consider $\limsup{a_{n+1}/a_n}$. First suppose it's larger than 1.
Since $(1+x)/y > x/y$ when x and y are positive, this means we can find $\epsilon>0$ and a subsequence for which $((1+a_{n+1})/a_n)^n > (1+\epsilon)^n$ and hence the limsup is infinity.

On the other hand if $\limsup{a_{n+1}/a_n} <1$ then the sequence decays at least as fast as a geometric sequence and must go to zero. As soon as $a_n<1$ we have $(1+a_{n+1})/a_n > 1/a_n >1$ and again the limsup must be infinity. So the only interesting thing is what happens when the limsup of this ratio is exactly 1.

Edit: now we can consider $\limsup{(1+a_{n+1})/a_n}$. This must be at least 1 since all we've done is add 1 to the numerator. If it's larger than 1 then clearly $\limsup{((1+a_{n+1})/a_n)^n}$ is infinity. So this must also be 1. For the limsup to be 1 in both cases it must be that $a_n\to \infty$ whenever we restrict to a subsequence that makes $a_{n_k+1}/a_{n_k}\to 1$.

So we are just interested in sequences $a_n$ going to infinity, and We can rewrite the original expression as $(1+(a_{n+1}-a_n+1)/a_n)^n$ and ,I guess the point is since we know the sequence is going to infinity, $(a_{n+1}-a_n+1)/a_n)$ is at least $1/n$. Intuitively this is because if $a_n$ grows slower than $n$, then we have at least $1/a_n$ and we're done. If $a_n$ grows faster, then the $(a_{n+1}-a_n)/a_n$ is at least $1/n$. Turning this into a formal proof seems tricky and I suspect I missed a clever step to avoid a lot of this.