Hey!!

I want to find for the following series the radius of convergence and the set of $x\in \mathbb{R}$ in which the series converges.

- $\displaystyle{\sum_{n=0}^{\infty}\frac{n}{2^n}x^{n^2}}$
- $\displaystyle{\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}}(x-1)^{3n}}$

I have done the following:

- $a_n=\frac{n}{2^n}$

$$|\frac{a_n}{a_{n+1}}|=\frac{n2^{n+1}}{(n+1)2^n}=\frac{2n}{n+1}=\frac{2}{1+\frac{1}{n}}$$

So, $$R=\lim_{n\rightarrow \infty}|\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow \infty}\frac{2}{1+\frac{1}{n}}=2$$

The radius of convergence is $2$. The series converges at $|x|<2$ and diverges at $|x|>2$.

For $x=\pm 2$ we have the following:

$$\sum_{n=0}^{\infty}\frac{n}{2^n}(\pm 2)^{n^2}$$

How could we continue?- $a_n=\frac{1}{(4+(-1)^n)^{3n}}$

$$\sqrt[n]{|a_n|}=\sqrt[n]{\frac{1}{(4+(-1)^n)^{3n}}}=\frac{1}{(4+(-1)^n)^3}$$

For $n=2k$ : $$\lim_{k\rightarrow \infty}\frac{1}{(4+(-1)^{2k})^3}=\frac{1}{5^3}$$

For $n=2k+1$ : $$\lim_{k\rightarrow \infty}\frac{1}{(4+(-1)^{2k+1})^3}=\frac{1}{3^3}$$

So, $\frac{1}{R}=\lim\sup\sqrt[n]{|a_n|}=\frac{1}{3^3}$.

The radius of convergence is $3^3=27$. The series converges at $|x|<27$ and diverges at $|x|>27$.

For $x=27$ we have the following:

$$\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}26^{3n}}$$ How could we check if the series converges?

For $x=-27$ we have the following:

$$\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}(-28)^{3n}}=\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}28^{3n}(-1)^n}$$ Is this an alternating series? Do we use here the Leibniz criterium?