Root Test For Convergence Question 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 $q&lt1$ then $\frac{1+q}{2}&lt1$ 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.
 A: 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 $a&lt1$ then $\frac{1}{a}>1$ and the result follows from the case $b>1$. 
A: 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 $q&lt1$, or 'blows up' if $q>1$.
By assumption, $\forall \epsilon >0$, $\exists n$ such that $|\sqrt[n]{a_n}-q| &lt \epsilon$. To use the hints you have provided, you just need to choose $\epsilon$ appropriately.
For Claim (1), you would like to have
$$\sqrt[n]{a_n} &lt \frac{1+q}{2}$$
Subtracting $q$ from both sides, we get
$$\sqrt[n]{a_n} -q  &lt \frac{1-q}{2}$$
Since $q&lt1$, the right hand side is positive, so choose $\epsilon = \frac{1-q}{2}$, in the limit definition. This gives
$$a_n &lt (\frac{1+q}{2})^n$$
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.
