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?


  • 1
    $\begingroup$ Although it's correct, I think step 4 as written is misleading. The idea is to exponentiate both sides of the equation in step 3, then to use the continuity of the exponential function to commute said function with the limit. Therefore I would not have put the limit after the log but before it, in the first equation of step 4. $\endgroup$
    – Simon
    Commented Aug 20, 2018 at 12:49

2 Answers 2


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.

  • $\begingroup$ See, this is what I don't understand. By this logic it turns out that the limit of a logarithm of some function is the same as the logarithm of the limit of that function. That doesn't sound to me like something that's generally correct. Or am I missing something? $\endgroup$
    – Koy
    Commented Aug 20, 2018 at 13:15
  • $\begingroup$ I meant sequence instead of function, sorry. $\endgroup$
    – Koy
    Commented Aug 20, 2018 at 13:18
  • 6
    $\begingroup$ @Koy : this is true not only for logarithm but for any continuous function. $\endgroup$
    – Paramanand Singh
    Commented Aug 20, 2018 at 13:46

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.


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