$R$ is called the radius of convergence of the series: if the series $\sum a_n (x-a)^n$ has radius of convergence $R$, then the series will converge when $|x-a| and diverge when $|x-a|>R$. One can calculate $R=\lim\limits_{n\rightarrow\infty} {|a_n|\over|a_{n+1}|}$, when this limit exists.
Your series does not have the required form to use the above directly. However, if you take the limit above, you'll obtain $\lim\limits_{n\rightarrow\infty} {|a_n|\over|a_{n+1}|}=3$. Then you can state that:
$\ \ \ $ The series converges absolutely when $|x+1|^5< 3$.
$\ \ \ $ The series diverges when $|x+1|^5>3$.
You need to solve the inequality $|x+1|^5<3$ to find the interior of the interval of convergence. Doing so gives $|x+1|^5<3$ if and only if $x$ is in the interval $( -{\root 5\of 3}-1, \root 5\of 3 -1 ) $.
At this point, we can say that the series converges for $x$ in this interval and that the series diverges for $x$ not in the interval $[ -{\root 5\of 3}-1, \root 5\of 3 -1 ] $ (this is precisely when |x+1|^5>3). Please note that we know nothing about the series when $x$ is an endpoint of this interval.
So:
$\ \ \ 1)$ The series converges for $x$ in $( -{\root 5\of 3}-1, \root 5\of 3 -1 ) $
$\ \ \ 2)$ The series diverges for $x$ not in $[ -{\root 5\of 3}-1, \root 5\of 3 -1 ] $.
But, at this point and as mentioned above, we do not know how the series behaves at the endpoints of the interval. We have to examine the series obtained when we set $x= -{\root 5\of 3}-1$ and $x= \root 5\of 3 -1 $.
When $x= -{\root 5\of 3}-1$, we have the series $ \sum{\log(n+12)\over (n+12)3^n}((-{\root 5\of 3} -1) +1)^{5n} =\sum{\log(n+12)\over (n+12)3^n}( - 3)^{ n} =\sum( - 1)^{ n}{\log(n+12)\over (n+12) } $ This is a convergent alternating series. So
$\ \ \ 3)$ The original series converges for $x= -{\root 5\of 3}-1$
When $x= {\root 5\of 3}-1$, we have the series $ \sum{\log(n+12)\over (n+12)3^n}(( {\root 5\of 3} -1) +1)^{5n} =\sum{\log(n+12)\over (n+12)3^n}( 3)^{ n} =\sum {\log(n+12)\over (n+12) } $ One can compare the series on the right above with the Harmonic series to show that it diverges.
So
$\ \ \ 4)$ The original series diverges for $x= {\root 5\of 3}-1$.
Summarizing $1)$ through $4)$:
The series converges if ind only if $x$ is in the interval $[ -{\root 5\of 3}-1, \root 5\of 3 -1 )$.
I think it's best to start the problem just using the Ratio test. That is, take the limit of
$|a_{n+1} x^{5(n+1)}\over |a_n (x+1)^{5n}|$ $ \lim_{n\rightarrow\infty} {\Bigl|{ \log((n+1)+12) (x+1)^{5(n+1)}\over ((n+1)+12)3^{n+1} }\Bigr| \over \Bigl|{ \log(n+12) (x+1)^{5n}\over (n+12)3^n }\Bigr|} =\cdots= \lim_{n\rightarrow\infty} {|x+1|^5\over3}. $ By the Ratio test, the series will converge when
$ {|x+1|^5\over3}<1$ and will diverge when
$ {|x+1|^5\over3}>1$ .
This tells you that the series will converge when $ {|x+1|^5 }<3$ and will diverge when $ {|x+1|^5 }>3$ .