Thursday, August 29, 2019

real analysis - Why is $limlimits_{n to +infty }{sqrt[n]{a_1 a_2 ldots a_n}} =limlimits_{n to +infty}{a_n}$



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 $e$.


Answer



Take the logarithm of both sides. Then you want to prove



$$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_1^n a_i = \lim_{n\rightarrow \infty} a_n.$$




This is a standard result about Cesàro means.


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