Analysis: Determine radius and interval of convergence for these power series I needed to determine the radius and interval of convergence for each of the following;
$\mathbb{One:}$ $$\sum_{n=0}^{\infty}\frac{3^nx^n}{n!}$$
$\mathbb{Two:}$ $$\sum_{n=0}^{\infty}\frac{x^n}{(1+n^2)}$$
$\mathbb{Three:}$ $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x+1)^n}{n}$$
 A: $\mathbb{One:}$ Radius of convergence is $\infty$, interval of convergence is $\mathbb{R}$. Apply the ratio test and calculate:
$$\lim_{{n}\rightarrow{\infty}}\left | \frac{3^{n+1}x^{n+1}/(n+1)!}{3^nx^n/n!}\right|=|x|\space \lim_{{n}\rightarrow{\infty}}\frac{3}{n+1}=0$$
Hence, this series converges absolutely for all values of $x$ (since this limit is $0$ for every value of $x$.
$\mathbb{Two:}$ Radius of convergence is $1$, interval of convergence is $[-1,1]$. Apply the ratio test and calculate:
$$\lim_{{n}\rightarrow{\infty}}\left | \frac{x^{n+1}/(1+(n+1)^2)}{x^n/(1+n^2)}\right|=|x|\space \lim_{{n}\rightarrow{\infty}}\frac{1+n^2}{2+2n+n^2}=\left | x \right |.$$
Hence, this series converges absolutely for $\left | x \right |<1$, and so the radius of converge is $1$. We now need to check the endpoints of the interval $(-1,1)$:
At $x=-1$, the series becomes $\sum_{n=0}^{\infty}\frac{(-1)^n}{1+n^2}$, which converges absolutely. 
At $x=1$, the series becomes $\sum_{n=0}^{\infty}\frac{1}{1+n^2}$, which converges absolutely. 
So, the series converges absolutely for all $x$ in the closed interval $[-1,1]$, and diverges elsewhere.
$\mathbb{Three:}$ Radius of convergence is $1$, interval of convergence is $(−2,0]$. Apply the ratio test and calculate:
$$\lim_{{n}\rightarrow{\infty}}\left | \frac{(-1)^{(n+1)+1}(x+1)^{n+1}/(n+1)}{(-1)^{n+1}(x+1)^n/n}\right|=|x+1|\space \lim_{{n}\rightarrow{\infty}}\frac{n}{n+1}=\left | x +1\right |.$$
Hence, this series converges absolutely for $|x + 1| < 1$, and so the radius of convergence is $1$. We now need to check the endpoints of the interval $(−2, 0)$:
At $x = −2$, the series becomes  $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-1)^n}{n}=-\sum_{n=1}^{\infty}\frac{1}{n}$, which diverges, being a constant multiple of the harmonic series.
At $x = 0$, the series becomes $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$, which converges conditionally, as it is the alternating harmonic series.
So, the series converges absolutely for all x in the open interval $(−2, 0)$, converges conditionally at $x = 0$, and diverges elsewhere.
