Solutions - 0220-01

(a)

$$\begin{array}{c}|a_{n+1}{x}^{n+1}\cdot a_n^{-1}{x}^{-n}|=|\frac{x^{n+1}}{{(n+1)}^2{3}^{n+1}}\cdot \frac{n^2{3}^n}{x^n}|\\ \\ \gg \gg \frac{n^2}{3{(n+1)}^2}|x|\gg \gg \frac{n^2}{3n^2+6n+3}|x|\stackrel{n\to \infty }{⟶}\frac{1}{3}|x|\end{array}$$

Therefore, the radius of convergence is $3$ and the preliminary interval is $(-3,+3)$.

Check end points:

$$\begin{array}{cc}x& =-3:\sum _{n=1}^{\infty }\frac{x^n}{n^2{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n{3}^n}{n^2{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}\\ & \\ x& =+3:\sum _{n=1}^{\infty }\frac{x^n}{n^2{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{3^n}{n^2{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{1}{n^2}\end{array}$$

Both of these series converge, so the final interval of convergence is $I=[-3,+3]$.

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(b)

$$\begin{array}{c}|a_{n+1}{x}^{n+1}\cdot a_n^{-1}{x}^{-n}|=|\frac{x^{n+1}}{(n+1)3^{n+1}}\cdot \frac{n{3}^n}{x^n}|\\ \\ \gg \gg \frac{n}{3(n+1)}|x|\gg \gg \frac{n}{3n+3}|x|\stackrel{n\to \infty }{⟶}\frac{1}{3}|x|\end{array}$$

Therefore, $R=3$ and the preliminary interval is $(-3,+3)$.

Check end points:

$$\begin{array}{cc}x& =-3:\sum _{n=1}^{\infty }\frac{x^n}{n{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n{3}^n}{n{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}\\ & \\ x& =+3:\sum _{n=1}^{\infty }\frac{x^n}{n{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{3^n}{n{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{1}{n}\end{array}$$

The first series converges by the AST. The second diverges ($p\ge 1$).

So the final interval of convergence is $I=[-3,+3)$.

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(c)

$$\begin{array}{c}|a_{n+1}{x}^{n+1}\cdot a_n^{-1}{x}^{-n}|=|\frac{x^{n+1}}{3^{n+1}}\cdot \frac{3^n}{x^n}|\gg \gg \frac{1}{3}|x|\end{array}$$

Therefore, the radius of convergence is $3$ and the preliminary interval is $(-3,+3)$.

Check end points:

$$\begin{array}{cc}x& =-3:\sum _{n=1}^{\infty }\frac{x^n}{3^n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n{3}^n}{3^n}\gg \gg \sum _{n=1}^{\infty }{(-1)}^n\\ & \\ x& =+3:\sum _{n=1}^{\infty }\frac{x^n}{3^n}\gg \gg \sum _{n=1}^{\infty }\frac{3^n}{3^n}\gg \gg \sum _{n=1}^{\infty }1\end{array}$$

Both series diverge. So the final interval is $I=(-3,+3)$.