Proving Boundedness of Seq. $(a_n)$ Given Bound Seq. $(\frac{a_{n+1}}{a_n})$

[mathmari]

Hey! :giggle:

Show for each sequence $(a_n)\subset (0, \infty)$ for which the sequence $\left (\frac{a_{n+1}}{a_n}\right )$ is bounded, that $\sqrt[n]{a_n}$ is also bounded and that $$\lim \sup \sqrt[n]{a_n}\leq \lim \sup \frac{a_{n+1}}{a_n}$$ I have done teh following:

The sequence $\left (\frac{a_{n+1}}{a_n}\right )$ is bounded. Let $L:=\limsup_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}\in \mathbb{R}$. Since all numbers $\frac{a_{n+1}}{a_n}$ are positive, it holds that $L \geq 0$.

We consider an arbitrary number $b$ with $0 \leq L < b$. It exists a $n_0\in \mathbb{N}$, such that $0<\frac{a_{n+1}}{a_n}<b$ for all $n\geq n_0$.

For all $n\geq n_0$ we have :
\begin{align*}0<\sqrt[n]{a_n}&=a_n^{\frac{1}{n}}=\left (a_n\cdot \frac{a_{n-1}}{a_{n-1}}\cdot \frac{a_{n-2}}{a_{n-2}}\cdot \frac{a_{n-3}}{a_{n-3}}\cdot \ldots \cdot \frac{a_{n_0+1}}{a_{n_0+1}}\cdot \frac{a_{n_0}}{a_{n_0}}\right )^{\frac{1}{n}} =\left ( \underset{(n-n_0)-\text{Factors}}{\underbrace{\frac{a_{n}}{a_{n-1}}\cdot \frac{a_{n-1}}{a_{n-2}}\cdot \frac{a_{n-2}}{a_{n-3}}\cdot \ldots \cdot \frac{a_{n_0+1}}{a_{n_0}}}}\cdot a_{n_0}\right )^{\frac{1}{n}}\\ & <\left ( b^{n-n_0}\cdot a_{n_0}\right )^{\frac{1}{n}}=\left ( b^{n}\cdot b^{-n_0}\cdot a_{n_0}\right )^{\frac{1}{n}}=b\cdot \left ( b^{-n_0}\cdot a_{n_0}\right )^{\frac{1}{n}}=b\cdot \sqrt[n]{ b^{-n_0}\cdot a_{n_0}}\end{align*}
For the fixed number $q:=b^{-n_0}\cdot a_{n_0}$ the sequence $(\sqrt[n]{q})$ converges to $1$, and so each subsequence converges to $1$.
This means that there is an $\epsilon>0$ such that $|\sqrt[n]{q}-1|<\epsilon \Rightarrow -\epsilon<\sqrt[n]{q}-1<\epsilon \Rightarrow 1-\epsilon<\sqrt[n]{q}<1+\epsilon $.

So $\sqrt[n]{a_n}$ is bounded above by $b(1+\epsilon)$.

Since $\sqrt[n]{a_n}<b\cdot \sqrt[n]{ b^{-n_0}\cdot a_{n_0}}$ the limit of each convergent subsequence of $(\sqrt[n]{a_n})$ is less or equal to $b$. So we get \begin{equation*}\limsup_{n\rightarrow \infty}\sqrt[n]{a_n}\leq b\end{equation*} This holds for all $b>\limsup_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=L$.
We choose $b=L+\frac{1}{m}$ for $m\in \mathbb{N}$.
Then \begin{equation*}\limsup_{n\rightarrow \infty}\sqrt[n]{a_n}\leq L+\frac{1}{m}, \ \ \ \text{ for all } m\in \mathbb{N}\end{equation*}
Taking the limit $m \rightarrow \infty$ we get \begin{equation*}\limsup_{n\rightarrow \infty}\sqrt[n]{a_n}\leq L \Rightarrow \limsup_{n\rightarrow \infty}\sqrt[n]{a_n}\leq\limsup_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}\end{equation*}

Is everything correct? Also the upper bound of the sequence $(\sqrt[n]{a_n})$ ?

:unsure:

 

[jvicens]

Hi there! Your proof looks good overall. However, I noticed a few minor errors that I wanted to point out:

1. In the beginning, you wrote "$L \geq 0$" but it should actually be "$L \geq 1$". This is because the sequence $(a_n)$ is bounded by $(0, \infty)$, so $a_{n+1} / a_n \geq 1$ for all $n$.

2. When you say "there is an $\epsilon > 0$ such that $|\sqrt[n]{q} - 1| < \epsilon \Rightarrow -\epsilon < \sqrt[n]{q} - 1 < \epsilon$", I think you meant to say "$\forall \epsilon > 0$ there exists an $N$ such that $\forall n > N$, $|\sqrt[n]{q} - 1| < \epsilon$". This is because you want to say that the sequence $(\sqrt[n]{q})$ converges to $1$, not just that there is one specific $\epsilon$ that works.

3. In the last step, when you take the limit $m \rightarrow \infty$, you can't just say that $\limsup_{n\rightarrow \infty}\sqrt[n]{a_n}\leq L + \frac{1}{m}$. Instead, you should say that $\limsup_{n\rightarrow \infty}\sqrt[n]{a_n} \leq L + \frac{1}{m}$ for all $m \in \mathbb{N}$. Then, you can take the limit as $m \rightarrow \infty$ and conclude that $\limsup_{n\rightarrow \infty}\sqrt[n]{a_n}\leq L$.

Other than those minor points, your proof looks solid. Good job!