Find the radius of convergence, R, of the series. n = 2 x7 +9 9n! Find the interval, I, of convergence of the series. (Enter your answer using interval notation.) Step 1 We will use the Ratio Test to determine the radius of convergence. We have the following. lim an+1 n→ ∞ lim n → ∞o an x + 10 9(n + 1)! +7 +9 9n! = lim x + 10 9 n! no 9(n+1)! n! x7 + 9 Step 2 Simplifying, we get lim n → ∞ n+1 Step 3 Therefore, we now have the following. lim n→ ∞ n+1 1 = |x| lim an noon + 1 = |×| · . X n+1
Find the radius of convergence, R, of the series. n = 2 x7 +9 9n! Find the interval, I, of convergence of the series. (Enter your answer using interval notation.) Step 1 We will use the Ratio Test to determine the radius of convergence. We have the following. lim an+1 n→ ∞ lim n → ∞o an x + 10 9(n + 1)! +7 +9 9n! = lim x + 10 9 n! no 9(n+1)! n! x7 + 9 Step 2 Simplifying, we get lim n → ∞ n+1 Step 3 Therefore, we now have the following. lim n→ ∞ n+1 1 = |x| lim an noon + 1 = |×| · . X n+1
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 50E
Question
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
Transcribed Image Text:Find the radius of convergence, R, of the series.
n = 2
x7 +9
9n!
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
Step 1
We will use the Ratio Test to determine the radius of convergence.
We have the following.
lim
an+1
n→ ∞
lim
n → ∞o an
x + 10
9(n + 1)!
+7 +9
9n!
=
lim
x + 10
9
n!
no 9(n+1)!
n!
x7 + 9
Step 2
Simplifying, we get lim
n → ∞
n+1
Step 3
Therefore, we now have the following.
lim
n→ ∞
n+1
1
=
|x| lim
an
noon + 1
= |×| ·
.
X
n+1
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