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$a_n$ is a sequence with $\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=L$, $a_n>0$.

So the task is to show that $c_n:=(a_n)^\frac{1}{n}$ converges and $\lim\limits_{n \rightarrow \infty}{c_n}=L$.

I've been working on this for hours now and don't have any useful result. Please help :)

namsap
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4 Answers4

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Let us assume $L > 0$ (otherwise, just use upper bounds).

Let $\epsilon > 0$. There exists $n_0 \in \Bbb N$ such that for every $n \geq n_0$ one has $L - \epsilon \leq \frac{a_{n+1}}{a_n} \leq L + \epsilon$. By an immediate induction this yields $$a_{n_0} (L-\epsilon)^n\leq a_{n+n_0} \leq a_{n_0}(L+\epsilon)^n$$ so $$ \frac{\log a_{n_0}}{n+n_0} + \frac{n}{n+n_0}\log (L-\epsilon) \leq \frac{\log a_{n+n_0}}{n+n_0} \leq \frac{\log a_{n_0}}{n+n_0} + \frac{n}{n+n_0}\log (L+\epsilon). $$ Letting $n\to+\infty$, we deduce that $$ \log(L-\epsilon) \leq \liminf \frac{\log a_n}{n} \leq \limsup \frac{\log a_n}{n} \leq \log(L+\epsilon) $$ Finally, let $\epsilon \to 0$ and take exponentials to conclude.

Siméon
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http://books.google.com.vn/books?id=WZX4GEpxPRgC&printsec=frontcover&hl=vi#v=onepage&q&f=false

This is a usual result, you can see the proof at Serge lang- Complex Analysis at page 57, in the aove link.

Hope it will help you.

An Khuong Doan
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  • Please try to describe as much here as possible in order to make the answer self-contained. Links are fine as support, but they can go stale and then an answer which is nothing more than a link loses its value. – robjohn Nov 24 '13 at 14:04
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We will show that $$\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=\lim_{n\to \infty} \sqrt[n] {a_n}$$.(Because the limits exist).

First we will show that $$\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=L\geq \lim_{n\to \infty} \sqrt[n] {a_n}$$.

Let $ε>0$ then there is a $n_1\in \Bbb N:\frac {a_{n+1}}{a_n}\leq L+\frac {ε}{2}$ for every $n\geq n_1$.

Now write $$a_n=\frac {a_n}{a_{n-1}}\cdot \frac {a_{n-1}}{a_{n-2}}\cdot ...\frac {a_{n_1+1}}{a_{n_1}}\cdot a_{n_1}$$. Thus we have that $$a_n\leq \frac {a_{n_1}}{(L+\frac {ε}{2})^{n_1}}\cdot (L+\frac {ε}{2})^n$$.

Let now $$θ=\frac {a_{n_1}}{(L+\frac {ε}{2})^{n_1}}$$ and by using that $\sqrt[n] {θ}\to 0$ we have that there is a $n_0\geq n_1:\sqrt [n] {a_n}\leq \sqrt[n] {θ}\cdot (L+\frac {ε}{2})<L+ε$ for every $n\geq n_0$ and thus $$\lim_{n\to \infty} \sqrt[n] {a_n}\leq L$$.

Same you do to prove $$\lim_{n\to \infty} \sqrt[n] {a_n}\geq L$$.

Haha
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Hint: For any $\epsilon\gt0$, there is an $N$ so that if $n\ge N$, $$ |\log(a_{n+1})-\log(a_n)-L|\le\epsilon $$ Thus, by the triangle inequality $$ \left|\ \left(\sum_{k=N}^{n-1}\log(a_{k+1})-\log(a_k)\right)-(n-N)L\ \right|\le(n-N)\,\epsilon $$ Therefore, for $n\ge N$, $$ \begin{align} &\left|\ \frac{\log(a_n)}{n}-L\ \right|\\ &\le\left|\ \frac{\log(a_n)-\log(a_0)}{n}-L\ \right|+\left|\ \frac{\log(a_0)}{n}\ \right|\\ &\le\left|\ \frac1n\left(\sum_{k=0}^{N-1}\log(a_{k+1})-\log(a_k)\right)\ \right|\\ &+\;\left|\ \frac1n\left(\sum_{k=N}^{n-1}\log(a_{k+1})-\log(a_k)\right)+\frac1n(n-N)L\ \right|\\ &+\;\frac1nNL\\ &+\;\left|\ \frac{\log(a_0)}{n}\ \right|\\[12pt] &\to\;0+\epsilon+0+0 \end{align} $$

robjohn
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