In a sequence of positive terms, $\lim \frac{a_{n+1}}{a_n} = L \Rightarrow \lim (a_n)^{\frac{1}{n}} = L$.

Question

Here's the question:

Let $(a_n)$ be a sequence of positive reals. Assume that $\lim \frac{a_{n+1}}{a_n} = L$. Then show that $\lim (a_n)^{\frac{1}{n}} = L$.

I tried everything including using the definition of convergence to prove it but nothing seems to work. There's no hint in the textbook as well. I'd appreciate if someone drops me some hint. I have been stuck in this problem for almost an hour!

Alternatively, use Cesàro summation on $u_n=\log\frac{a_n}{a_{n-1}}$. — Jean-Claude Arbaut, May 29 '18 at 05:26

score -6 · Answer 1 · edited May 29 '18 at 06:20

-6

try this: let $r_1=a_1$ and $r_n= \frac{a_{n}}{a_{n-1}}$ for $n>1$. Then the condition says $\lim_{n\to\infty} r_n = L$. Next express $a_n = r_1...r_n$. This should work well.

edited May 29 '18 at 06:20

Riemann

7,203

answered May 29 '18 at 05:28

user565744

1

In a sequence of positive terms, $\lim \frac{a_{n+1}}{a_n} = L \Rightarrow \lim (a_n)^{\frac{1}{n}} = L$.

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