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$a_n$ is sequence of positive reals. given $\lim \frac{a_{n+1}}{a_n} = l$. prove $\lim(a_n)^\frac{1}{n}=l$

For a given $\epsilon$, there exists a $N$ such that for all natural numbers greater than or equal to N following happens :

$|\frac{a_{n+1}}{a_n}- l| < \epsilon$

$\frac{a_{n+1}}{a_n} < l + \epsilon$

$a_{n+1} < a_n(l+\epsilon)$

$a_{N+1} < a_N(l+\epsilon)$

$a_{N+2} < (a_N)_{N+1}(l+\epsilon)$ ....

By Iteration

$a_{N+k} < a_N (l + \epsilon)^k$

This is as far i can reach. Thanks for help

Sophie Clad
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  • The right hand limit isn't even defined when $a_n$ is allowed to be negative, so presumably you have assumption that the sequence is positive. – Thomas Andrews Apr 06 '23 at 14:04
  • The probably backwards approach is to use the ratio test and root test for the radius of convergence of $f(x)=\sum_{n} a_nx^n.$ The ratio test says the radius of convergence is $1/l.$ The root test says this means $\limsup_{n\to\infty}a_n^{1/n}=l.$ So you "only" need to prove that $\liminf_{n\to\infty} a_n^{1/n}=l.$ (I say backwards because the proof probably wants to justify this equality outside those two theorems.) – Thomas Andrews Apr 06 '23 at 14:08
  • See https://math.stackexchange.com/q/69386/42969 or https://math.stackexchange.com/q/472562/42969 – Martin R Apr 06 '23 at 14:24
  • @ThomasAndrews Yes they are positive – Sophie Clad Apr 06 '23 at 14:30
  • Put it in the question, @SophieClad Answerers shouldn't have to read comments to know the full question. – Thomas Andrews Apr 06 '23 at 14:33
  • @ThomasAndrews how to know where to stop iteration ? i mean from $n \geq N$, how many values to put ? – Sophie Clad Apr 06 '23 at 14:36

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