Let $\epsilon>0$. There is an $N$ so that for all $n\ge N$, $|a_{n+1}|\le (r+\epsilon)|a_n|$
From this it follos that
$\ \ \ |a_{N+1}|\le (r+\epsilon)|a_N|$
$\ \ \ |a_{N+2}|\le (r+\epsilon)^2|a_N|$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$
$\ \ \ |a_{N+k}|\le (r+\epsilon)^k|a_N|$
Using the above, if $n>N$:
$$
|a_n|=|a_{N+(n-N)}|\le (r+\epsilon)^{n-N}|a_N|
$$
Let $A= |a_N|/(r+\epsilon)^N$.
Then
$$
\root n\of {|a_n| }\le A^{1/n}(r+\epsilon)
$$
for all $n>N$.
Now, $A^{1/n}\rightarrow1$ as $n\rightarrow\infty$; so,
$$
\limsup_{n\rightarrow\infty} \root n\of{|a_n|}\le r+\epsilon.
$$
Since $\epsilon$ was arbitrary, we have
$$
\limsup_{n\rightarrow\infty} \root n\of{|a_n|}\le r.
$$
Now show that $\liminf\limits_{n\rightarrow\infty} \root n\of{|a_n|}\ge r$. I'll leave that for you.
Note that the above can be modified slightly to show that $\limsup\limits_{n\rightarrow\infty}\root n\of {|a_n|}\le\limsup\limits_{n\rightarrow\infty}\root n\of {\Bigl|{a_{n+1}\over a_n}\Bigr|}$. One can also show that $\liminf\limits_{n\rightarrow\infty}\root n\of {|a_n|}\ge\liminf\limits_{n\rightarrow\infty}\root n\of {\Bigl|{a_{n+1}\over a_n}\Bigr|}$.
From this, your result easily follows.
Note also please, that the result has little to do with series...