For which $x \in \mathbb{R}$ does the following series converge? $$\sum_{n=1}^\infty \frac{3x^n}{2+x^{4n}}$$
I think the solution is $|x| \neq 1$
But I don't really know how to prove it rigorously.
Thanks in advance :)
 A: Low-tech solution: I would divide into cases according to whether $|x|$ is smaller than, larger than, or equal to $1$.
For $|x|<1$, the series is absolutely less than $\sum x^n$.
For $|x|>1$, we have that $2+x^{4n} > x^{4n}$ so the series is absolutely less than $\sum 3x^{-3n}$, which is a geometric series in $x^{-3}$.
Finally, for $|x|=1$ all terms are either $1$ or $-1$.
A: If $|x| > 1$ then:
$$\sum_{n=1}^\infty \frac{3|x|^n}{2+x^{4n}} \le \sum_{n=1}^\infty \frac{3|x|^n}{x^{4n}} = \sum_{n=1}^\infty \frac{1}{|x|^{3n}} = \sum_{n=1}^\infty \left(\frac{1}{|x|^3}\right)^n$$
which converges so the series converges absolutely.
If $|x| < 1$ then:
$$\sum_{n=1}^\infty \frac{3|x|^n}{2+x^{4n}} \le \sum_{n=1}^\infty 3|x|^n = 3 \sum_{n=1}^\infty |x|^n$$
which converges so the series converges absolutely.
If $|x| = 1$ then:
$$\sum_{n=1}^\infty \frac{3x^n}{2+x^{4n}} = \sum_{n=1}^\infty \frac{3x^n}3 = \sum_{n=1}^\infty {x^n}$$
which diverges.
A: If $x=1$, it diverges.
If $x=-1$, it diverges.
If $|x|<1$ then
$$u_n (x)\sim \frac {3}{2}x^n $$
thus it converges.
If $|x|>1$ then
$$u_n (x)\sim \frac {3}{x^{3n}} $$
thus it converges.
A: Let $a_n=\dfrac {3x^n}{2+x^{4n}}$ all cases can be dealt with comparison test.


*

*If $|x|<1$ then $|a_n|\sim \frac 32|x|^n$ and $\sum a_n$ converges absolutely.

*If $x=1$ then $a_n=1$ and $\sum a_n$ diverge.

*If $x=-1$ then $a_n=(-1)^n$ and $\sum a_n$ diverge as well.

*If $|x|>1$ then $|a_n|\sim \dfrac{3|x|^n}{|x|^{4n}}\sim\dfrac 3{|x|^{3n}}$ and $\sum a_n$ converges absolutely.
Thus the series converges absolutely for all $x\neq 1,-1$ and it diverges for these values.
