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While reading the proof of the root test for convergence in my notebook I came across this two claims:

let $\{a_n\}$ be a positive sequence and $\lim \limits_{n\to \infty}\sqrt[n]{a_n}=q. $

claim (1): if $q<1$ then $\frac{1+q}{2}<1$ therefore $\lim \limits_{n\to \infty}(\frac{1+q}{2})^n=0$ - how do I prove that the sequence converges to $0$?

claim (2): if $q>1$ then $\frac{1+q}{2}>1$ therefore $\lim \limits_{n\to \infty}(\frac{1+q}{2})^n=\infty$ - how do I prove that the sequence converges to $\infty$?

Thanks a lot.

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up vote 3 down vote accepted

If $a>1$ then $a=1+h$ for some $h>0$. By the binomial theorem $a^n=(1+h)^n\geq 1+nh$ for $n\geq2$. It follows easily from the last inequality that $a^n\to\infty.$ If $a<1$ then $\frac{1}{a}>1$ and the result follows from the case $b>1$.

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I understood the the first case where $a>1$; what did you mean afterwards when you wrote that the result follows from the case $b>1$? I didn't quite understand how to show that if $a<1$ then the sequence converges to 0. thanks a lot. – Anonymous Apr 12 '12 at 21:44
@Anonymous If $b_n\to \infty$ then $\frac{1}{b_n}\to 0$. – azarel Apr 12 '12 at 21:54

The idea is that since $\lim \limits_{n\to \infty}\sqrt[n]{a_n}=q$, and $n$ is large, then $a_n$ behaves like $q^n$, and so converges to zero if $q<1$, or 'blows up' if $q>1$.

By assumption, $\forall \epsilon >0$, $\exists n$ such that $|\sqrt[n]{a_n}-q| < \epsilon$. To use the hints you have provided, you just need to choose $\epsilon$ appropriately.

For Claim (1), you would like to have $$\sqrt[n]{a_n} < \frac{1+q}{2}$$ Subtracting $q$ from both sides, we get $$\sqrt[n]{a_n} -q < \frac{1-q}{2}$$ Since $q<1$, the right hand side is positive, so choose $\epsilon = \frac{1-q}{2}$, in the limit definition. This gives $$a_n < (\frac{1+q}{2})^n$$ for $n$ sufficiently large, and the comparison test shows that $a_n$ converges to $0$.

To prove the other claim, you repeat the same procedure, starting from the 'opposite' inequality.

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