Radius of convergence of $\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n$ I want to determine the convergence of the following series in dependency of $x$:
$\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n=\frac{3}{4}x+\frac{2}{5}x^2+\frac{1}{4}x^3+\frac{3}{17}x^4+ ... $
How can I solve this?
EDIT:
@Winther said, I should try the ratio test:
$q := \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n}\right| $
So we get
$q = \lim_{n \to \infty} \left| \frac{\frac{n+3}{2(n+1)^2+2}}{\frac{n+2}{2n^2+2}}\right| =  \lim_{n \to \infty} \left| \frac{n+3}{2(n+1)^2+2}\frac{2n^2+2}{n+2} \right| = \lim_{n \to \infty} \left| \frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}\right| = \lim_{n \to \infty} \left| \frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}\right| $
With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:
$ q = \lim_{n \to \infty} \left| \frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}\right|= 1$
So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).
Should I try another test (e. g. the root test)?
EDIT 2: Corrected the first coefficients.

Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
\to 1$).
Therefore
$f(x)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$\sum\limits_{n=1}^{\infty} x^n$
which is
$-1 < x < -1$.
At $x=1$,
the series is
$f(1)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} 
$
which diverges like the
harmonic series.
At $x=-1$,
the series is
$f(-1)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} (-1)^n
$
which converges 
because it is
an alternating series
with decreasing terms.
