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.
Best Answer
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.