Solved: "Find the radius of convergence and interval of..."

allhvasstH

Answered question

2020-11-08

Find the radius of convergence and interval of convergence of the series.

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 10^{n} x^{n}}{n^{4}}

Answer & Explanation

cyhuddwyr9

Skilled2020-11-09Added 90 answers

Here $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 10^{n} x^{n}}{n^{4}} = \sum_{n = 1}^{\infty} a_{n}, w h e r e a_{n} = \frac{(- 1)^{n} 10^{n} x^{n}}{n^{4}}$
Now $| \frac{a_{n} + 1}{a_{n}} | = | \frac{10^{n + 1} x^{n + 1} n^{2}}{10^{n} x^{n} {(n + 1)}^{2}} |$
$= | 10 x {(\frac{n}{n + 1})}^{2} |$
This shows that
$lim_{n \to \infty} | \frac{a_{n} + 1}{a_{n}} | = | 10 x | lim_{n \to \infty} {(\frac{n}{n + 1})}^{2} = 10 | x |$
Therefore the given series converges if
$10 | x | < 1 \Rightarrow | x | < \frac{1}{10}$
and diverges if $| x | > \frac{1}{10}$ . Therefore the radious of converges is $R = \frac{1}{10}$ .When $x = \frac{1}{10}$ the the series beomes $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{4}}$ which converges by the Leibniz's test. Again when $x = - \frac{1}{10}$ then the series becomes $\sum_{n = 1}^{\infty} \frac{1}{n^{4}}$ which converges by the p series test. Thereore the series converges converges in the interval $| x | \leq \frac{1}{10}$

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