small tip on If $a_n>0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l$ prove that $\lim a_n^{\frac{1}{n}}=l$

Question

I have the following problem:

If $a_n>0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l$ prove that $\lim_{n\to\infty} a_n^{\frac{1}{n}}=l$.

Now I know I have no work to show but I really don't even know how to start. I need a small tip on how to get going with this problem and not a whole solution.

Thanks in advance!

Seems to be answered here https://math.stackexchange.com/questions/418775/lim-n-rightarrow-infty-fraca-n1a-n-l-implies-lim-n-rightarrow-i?rq=1 — windircurse, Sep 22 '17 at 09:44

score 5 · Accepted Answer · answered Sep 22 '17 at 09:44

5

By Stolz-Cesaro theorem, $$\lim_{n\to\infty} \ln(a_n^{\frac{1}{n}})=\lim_{n\to\infty} \frac{\ln(a_n)}{n}=\lim_{n\to\infty} \frac{\ln(a_{n+1})-\ln(a_n)}{(n+1)-n}=\lim_{n\to\infty}\ln\left(\frac{a_{n+1}}{a_n}\right)=\ln(l).$$

answered Sep 22 '17 at 09:44

Robert Z

145,942

score 1 · Answer 2 · answered Sep 22 '17 at 15:29

Let $v_{1}=u_{1}, v_{2}=\frac{u_{2}}{u_1}, .....,v_{n}=\frac{u_{n}}{u_{n-1}},.....$

Then $v_{n}> 0$ for all $n\in \mathbb{N}$ and $\lim v_{n}=l>0.$

This imply $\lim \log v_{n}=\log l.$ Then by cauchy theorem,

$\lim \frac{\log{v_{1}}+\log{v_{2}}+......+\log{v_{n}}}{n}=\log l.$

i.e $\lim \log (v_{1}. v_{2}....v_{n})^{1/n}=\log l$.

i.e $ \lim (v_{1}. v_{2}....v_{n})^{1/n}=l.$

i.e $\lim u_{n}^{1/n}=l$.

small tip on If $a_n>0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l$ prove that $\lim a_n^{\frac{1}{n}}=l$

2 Answers2