Find the radius of convergence (-1)"x2n+1(2n + 1)!n=0

To find the radius of convergence for the series ∑n= 0∞-1nx2n + 12n + 1! ...(1)…

Answered: (-1)"x2n+1 (2n + 1)! n=0

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section: Chapter Questions

Problem 43RE

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Question

Find the radius of convergence

Expert Solution

Step 1

To find the radius of convergence for the series

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

Power series is given by $\sum_{n = 0}^{\infty} a_{n} {(x - a)}^{n}$

Comparing (1) with power series we get

$a_{n} = \frac{{(- 1)}^{n}}{(2 n + 1)!}$

Step 2

Formula:

Ratio Test:

$R a d i u s o f c o n v e r g e n c e o f a p o w e r s e r i e s c a n b e f o u n d b y \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = R R a d i u s o f c o n v e r g e n c e = \frac{1}{R}$

To find R:

$\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = \lim_{n \to \infty} |\frac{\frac{{(- 1)}^{n + 1}}{(2 (n + 1) + 1)!}}{\frac{{(- 1)}^{n}}{(2 n + 1)!}}| = \lim_{n \to \infty} |\frac{\frac{{(- 1)}^{n + 1}}{(2 n + 3)!}}{\frac{{(- 1)}^{n}}{(2 n + 1)!}}| = \lim_{n \to \infty} |\frac{{(- 1)}^{n + 1}}{(2 n + 3)!} . \frac{(2 n + 1)!}{{(- 1)}^{n}}| = \lim_{n \to \infty} |\frac{{(- 1)}^{n + 1}}{(2 n + 3) (2 n + 2) (2 n + 1)!} . \frac{(2 n + 1)!}{{(- 1)}^{n}}|$

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