Calculus II: Assignment 10 Solutions
Course: MAT 202 - Calculus II
Topic: Infinite Series, Convergence Tests, and Power Series
This document provides a detailed step-by-step walkthrough of the solutions for Assignment 10. The problems focus on determining the convergence of infinite series using various tests and analyzing the interval of convergence for power series.
Problem 1: Ratio Test for Convergence
Question: Determine whether the following series converges or diverges:
∑ (n=1 to ∞) (n! / 2n)
Solution
To determine the convergence of this series involving factorials and exponentials, the Ratio Test is the most appropriate method. The Ratio Test states that for a series ∑ an, we calculate the limit:
L = lim (n→∞) | an+1 / an |
If L < 1, the series converges absolutely. If L > 1, the series diverges. If L = 1, the test is inconclusive.
an = n! / 2n
an+1 = (n+1)! / 2n+1
an+1 / an = [ (n+1)! / 2n+1 ] ÷ [ n! / 2n ]
= [ (n+1)! / 2n+1 ] × [ 2n / n! ]
= [ (n+1) × n! / (2 × 2n) ] × [ 2n / n! ]
= (n+1) / 2
L = lim (n→∞) (n+1) / 2 = ∞
Since L = ∞, which is strictly greater than 1, the series diverges by the Ratio Test.
Problem 2: Interval of Convergence
Question: Find the radius of convergence and the interval of convergence for the following power series:
∑ (n=1 to ∞) ((-1)n (x + 2)n / n √n)
Solution
We will use the Ratio Test to find the radius of convergence (R). The inequality |an+1 / an| < 1 will allow us to solve for x.
an = (-1)n (x + 2)n / n √n
|an+1 / an| = | [ (-1)n+1 (x + 2)n+1 / (n+1) √(n+1) ] ÷ [ (-1)n (x + 2)n / n √n ] |
= |(-1)(x + 2)| × [ n √n / (n+1) √(n+1) ]
= |x + 2| × ( n / (n+1) )3/2
lim (n→∞) ( n / (n+1) )3/2 = 13/2 = 1
So, the limit expression is simply |x + 2|.
|x + 2| < 1
The inequality |x - c| < R implies a Radius of Convergence R = 1.
The inequality -1 < x + 2 < 1 gives -3 < x < -1.
This is the preliminary interval (-3, -1). We must check the endpoints.
x = -3: Series becomes ∑ ((-1)n (-1)n / n √n) = ∑ (1 / n3/2). This is a convergent p-series (p = 3/2 > 1). Include -3.
x = -1: Series becomes ∑ ((-1)n (1)n / n √n) = ∑ ((-1)n / n3/2). This converges absolutely. Include -1.
Interval of Convergence: [-3, -1]
Problem 3: Taylor Series Expansion
Question: Find the Maclaurin series (Taylor series centered at 0) for the function f(x) = x cos(x) and determine its radius of convergence.
Solution
Instead of calculating derivatives manually, we can utilize the known Maclaurin series for the cosine function.
cos(x) = ∑ (n=0 to ∞) (-1)n x2n / (2n)!
= 1 - x2/2! + x4/4! - x6/6! + ...
x · cos(x) = x · [ ∑ (n=0 to ∞) (-1)n x2n / (2n)! ]
= ∑ (n=0 to ∞) (-1)n x2n+1 / (2n)!
x - x3/2! + x5/4! - x7/6! + ...
The series for cos(x) converges for all real numbers (R = ∞). Multiplying by x does not change the domain of convergence.
Radius of Convergence: ∞
Note: These solutions are intended as a guide to demonstrate the methods required to solve Calculus II problems. Always ensure you understand the underlying concepts before applying these formulas on exams.
