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...