Explained: "Determine the radius of convergence of this series...." - Plainmath

sibuzwaW

Answered question

2020-12-17

Determine the radius of convergence of this series.

\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}

Answer & Explanation

liingliing8

Skilled2020-12-18Added 95 answers

The given series is $\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$
By the series ratio test, a series $\sum_{n = 0}^{\infty} a_{n}$ converges if $lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | < 1$
Compare the series $\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$ with $\sum_{n = 0}^{\infty} a_{n}$ and obtain $a_{n} = (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$
Now, $a_{n} = (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$ implies that $a_{n + 1} = (- 1)^{n + 1} \frac{x^{2 (n + 1) + 1}}{2 (n + 1) + 1} = (- 1)^{n + 1} \frac{x^{2 n + 3}}{2 n + 3}$
Therefore, by the ratio test, the series $\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$ converges if $lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} | \frac{(- 1)^{n + 1} \frac{x^{2 n + 3}}{2 n + 3}}{(- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}} | < 1$
Note that,
$lim_{n \to \infty} | \frac{(- 1)^{n + 1} \frac{x^{2 n + 3}}{2 n + 3}}{(- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}} | < 1$
$lim_{n \to \infty} | - \frac{(2 n + 1) x^{2 n + 3}}{(2 n + 3) x^{2 n + 1}} | < 1$
$\Rightarrow lim_{n \to \infty} | - \frac{n (2 + \frac{1}{n}) x^{2}}{n (2 + \frac{3}{n})} | < 1$
$\Rightarrow lim_{n \to \infty} | - \frac{(2 + \frac{1}{n}) x^{2}}{(2 + \frac{3}{n})} | < 1$
$\Rightarrow | - \frac{(2 + \frac{1}{\infty}) x^{2}}{(2 + \frac{3}{\infty})} | < 1$
$\Rightarrow | - \frac{2 x^{2}}{2} | < 1$
$\Rightarrow | - x^{2} | < 1$
$\Rightarrow | x^{2} | < 1$
$\Rightarrow {| x |}^{2} < 1$
$\Rightarrow | x | < 1$
Since the value $| x | < 1$ , the radius of convergence R of the series $\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$ is R=1

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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