Question

# Radius and interval of convergence Determine the radius and interval of convergence of the following power series. x-frac{x^3}{4}+frac{x^5}{9}-frac{x^7}{16}+...

Series
Radius and interval of convergence Determine the radius and interval of convergence of the following power series.
$$x-\frac{x^3}{4}+\frac{x^5}{9}-\frac{x^7}{16}+...$$

2020-10-24

First write the given series in formula:
$$S=x-\frac{x^3}{4}+\frac{x^5}{9}-\frac{x^7}{16}+...$$
$$S=x-\frac{x^3}{2^2}+\frac{x^5}{3^3}-\frac{x^7}{4^4}+...$$
$$S=\frac{x^{2(1)-1}}{1^2}-\frac{x^{2(2)-1}}{2^2}+\frac{x^{2(3)-1}}{3^2}-\frac{x^{2(4)-1}}{4^2}+...$$
$$S=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}x^{2n-1}$$
Apply ratio test and check weather series is convergent or divergent.
If $$|\frac{a_{n+1}}{a_n}|\leq q$$ eventually for some 0, then $$\sum_{n=1}^\infty|a_n|$$ converges.
If $$|\frac{a_{n+1}}{a_n}|>1$$ eventually then $$\sum_{n=1}^\infty a_n$$ diverges
Consider the given series:
$$f(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}x^{2n-1}$$
$$a_k=\frac{(-1)^{k+1}}{k^2}2^{2k-1}$$
$$a_{k+1}=\frac{(-1)^{k+2}}{(k+1)^2}x^{2k+2-1}$$
$$=\frac{(-1)^{k+2}}{(k+1)^2}x^{2k+1}$$
$$\frac{a_{k+1}}{a_k}=\frac{\frac{(-1)^{k+2}}{(k+1)^2}x^{2k+1}}{\frac{(-1)^{k+1}}{k^2}x^{2k-1}}$$
$$=\frac{(-1)^{k+2-k-1}k^2}{(k+1)^2}x^{2k+1-2k+1}$$
$$=\frac{(-1)k^2}{(k+1)^2}x^2$$
$$L=\lim_{k\to\infty}|\frac{a_{k+1}}{a_k}|$$
$$=\lim_{k\to\infty}|\frac{(-1)k^2}{(k+1)^2}x^2|$$
$$=|-x^2|$$
From ratio test, if $$L > 1$$ series diverges.
L<1 series converges.
$$|x|<1\Rightarrow-1\leq x\leq1$$ series converges.
Radius of convergence, $$R=1$$
Find the interval of convergence.
$$-1\leq x\leq1$$
Interval of convergence $$=\left\{x:x=-1\leq x\leq1\right\}$$
Interval of convergence $$[-1,1]$$