# Why is $\lim_{n \to +\infty }{\sqrt[n]{a_1 a_2 \cdots a_n}} =\lim_{n \to +\infty}{a_n}$

Solving some problems regarding limits and sequence convergence, i stumbled upon a task, and it's solution relies on, and i quote: "We now use a well-known theorem : $$\lim_{n \to +\infty }{\sqrt[n]{a_1 a_2 \ldots a_n}} = \lim_{n \to +\infty}{a_n}$$

This isn't really intuitive (at least to me) and I don't know how to prove it. The original task was to find the limit of $$\lim_{n \to +\infty }{\sqrt[n]{\bigg{(}1+\frac{1}{1}\bigg{)} \bigg{(}1+\frac{1}{2}\bigg{)}^2 \ldots \bigg{(}1+\frac{1}{n}\bigg{)}^n}}$$ which of course, using the expression above is just $e$.

• may I ask you which book you found this task in? I'm asking because it appeared among our exercise tasks and I still don't know how to prove this theorem (if it really is a theorem). It would really help me because I'm a first-year student and there are a very few books available in my mother-tongue. Dec 21, 2019 at 8:33
• The hypothesis $a_n>0$ should be stated.
– zhw.
Jan 5, 2020 at 17:52
• This question is a possible duplicate of yours. May 28, 2020 at 0:16
• You need to assume more than just $a_n\gt0$ for all $n$. For example, if $$a_n= \begin{cases} 1&\text{if n is odd}\\ 2&\text{if n is even} \end{cases}$$ then $\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=\sqrt2$ but $\lim_{n\to\infty}a_n$ does not exist. May 28, 2020 at 0:48

$$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_1^n a_i = \lim_{n\rightarrow \infty} a_n.$$
Take the log of the $n$-root, and applied the Cesaro theorem to it, showing that it will converge to the log of $(a_n)_n$'s limit. Take the exp to finish.