<s>[INST] 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.
 [/INST] If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $x\ne0$,
$$
\lim_{n\to\infty}
\left|\,\frac{\dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{\dfrac{n+2}{2n^2+2}x^n}\,\right|
=\lim_{n\to\infty}\frac{n+3}{n+2}\frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.
Thus the radius of convergence is $1$.
There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
\frac{1}{R}=\limsup_{n\to\infty}\sqrt[n]{a_n}
$$
but this can be difficult to compute.
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