Studying a convergence of a series of powers I'm studying for a calcus exam and I'm trying to solve some of the proposed exercises.
I need to study the convergence for $|x+1|^5=R$
$$\sum_{n=0}^{+\infty} \frac{\log(n+12)}{(n+12)\cdot 3^n}\cdot (x+1)^{5n}.$$
The limit of $a_n/a_{n+1}$ (using D'Alembert formula) gives me $R$
$$\lim_{n\to\infty} \frac{\log(n+12)}{(n+12)\cdot 3^n})\cdot(x+1)^{5n}\left(\frac{\log(n+1+12)}{(n+1+12)\cdot 3^{n+1}}\cdot(x+1)^{5(n+1)}\right)^{—1} 
= 3/(1+x)^5$$
But now I'm not sure what I need to do in order to study the convergence and I'd love a few pointers.
 A: $R$ 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 $|x-a|<R$ 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$ . 
A: Since $x$ is real, if $|x+1|^5=R=3$ then $(x+1)^5=\pm 3$. If $(x+1)^5=3$ then we have to study the convergence of $\sum_{n=0}^{+\infty}\frac{\log(n+12)}{n+12}$ and if $(x+1)^5=-3$ then we have to study the convergence of $\sum_{n=0}^{+\infty}(-1)^n\frac{\log(n+12)}{n+12}$. 
For the first, what do you know about the harmonic series? For the second you can try an Abel transform (summation by parts).
