Series area of convergence sadly I don't have any solutions. Is this correct?
$$\sum_{n=1}^\infty\frac{ 2^nx^n}{\sqrt{n^4+1}}$$
We calculate the radius of convergence:
$R=\lim_{x \to\infty}\big|\frac{a_n}{a_{ń+1}}\big|=\lim_{n\to\infty}\frac{1}{2}\sqrt{\frac{(n+1)^4+1}{n^4+1}}=\lim_{n\to\infty}\frac{1}{2}\sqrt{\frac{(1+1/n)^4+1/n^4}{1+1/n^4}}=1/2$
We check the boundary:
$x=\frac{1}{2}$
$$\sum_{n=1}^\infty\frac{2^n(\frac{1}{2})^n}{\sqrt{n^4+1}}=\sum_{n=1}^\infty\frac{1}{\sqrt{n^4+1}}$$
By comparing this to the harmonic series, we conclude that it does not converge.
$x=\frac{-1}{2}$
We get
$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n^4+1}}$$
Since $\frac{1}{\sqrt{n^4+1}}\geq 0 \quad \forall n$ and $\lim_{n\to\infty} \frac{1}{\sqrt{n^4+1}} = 0$ and because $\frac{1}{\sqrt{n^4+1}}$ is monotonic decreasing (since the square root is monotonic increasing) we can conclude, using Leibniz-Test, that this series does converge.
So we get: $-\frac{1}{2}\leq x < \frac{1}{2}$

The implication is not true for $x={1\over 2}$. By comparison test we have $$\sum_{n=1}^\infty\frac{1}{\sqrt{n^4+1}}<\sum_{n=1}^\infty\frac{1}{\sqrt{n^4}}=\sum_{n=1}^\infty\frac{1}{{n^2}}={\pi^2\over 6}$$
