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}+...

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asked 2020-10-23
Radius and interval of convergence Determine the radius and interval of convergence of the following power series.
\(x-\frac{x^3}{4}+\frac{x^5}{9}-\frac{x^7}{16}+...\)

Answers (1)

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 \([-1,1]\)

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