If $\lim_{n \to \infty} \frac{x_{n+1}}{x_n} = a > 0$, prove that $\lim_{n \to \infty} \sqrt[n]{x_n} = a$

Question

Sorry for the unclear title, the problem is too specific so I couldn't think of anything else. Here goes:

If $$ \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = a > 0, $$ prove that $$ \lim_{n \to \infty} \sqrt[n]{x_n} = a. $$

Now, in my textbook there is a proof provided but I don't understand it. It goes like this:

\begin{equation} \tag{1} \sqrt[n]{x_n} = \sqrt[n]{ \frac{x_n}{x_{n-1}} \times \frac{x_{n-1}}{x_{n-2}} \times \cdots \times \frac{x_2}{x_1} \times \frac{x_1}{1} } \end{equation}

Then they take $\log$ of both sides:

\begin{equation} \tag{2} \log \sqrt[n]{x_n} = \frac{1}{n} \left( \log \frac{x_n}{x_{n-1}} + \log \frac{x_{n-1}}{x_{n-2}} + \dotsb + \log \frac{x_2}{x_1} + \log \frac{x_1}{1} \right) \end{equation}

These two steps are clear to me. What comes next is what I don't understand:

\begin{equation} \tag{3} \lim_{n \to \infty} \log \sqrt[n]{x_n} = \log \lim_{n \to \infty} \frac{x_n}{x_{n-1}} = \log a \end{equation}

The textbook provides no other explanation for this except for a little note saying “Cauchy's theorem”. The only Cauchy theorem previously mentioned in the textbook was the first one in this article.

Then the rest of the proof looks like this:

\begin{equation} \tag{4} e^{\log \lim_{n \to \infty} \sqrt[n]{x_n}} = e^{\log a} \end{equation} \begin{equation} \tag{5} \lim_{n \to \infty} \sqrt[n]{x_n} = a \end{equation}

Where I also have no idea what's happening.

Any ideas?

Thanks.

Answer 1

I suppose that there is a typo in your textbook. When it mentions Cauchy, it should mention Cesàro summation instead, since it implies that\begin{align}\lim_{n\to\infty}\frac{\log\left(\frac{x_n}{x_{n-1}}\right)+\log\left(\frac{x_{n-1}}{x_{n-2}}\right)+\cdots+\log\left(\frac{x_1}{x_0}\right)}n&=\lim_{n\to\infty}\log\left(\frac{x_n}{x_{n-1}}\right)\\&=\log a.\end{align}By the way, it's a nice proof.

Answer 2

What's happenning is that if a sequence $a_n$ tends to a limit, then the sequence of averages $\frac{1}{n}\sum_{k=1}^na_k$ also tends to the same limit. This is Cauchy's first theorem of limits, according to the article you referred to.