a.
To find: To find whether the given series converges or diverges.
a.

Answer to Problem 1QQ
The series converges.
Explanation of Solution
Given:
The given series ∞∑n=N2n2+1 .
Formula used:
a=f(n)where f is a continuous, positive, decreasing function of x for all x≥N (N is a positive integer).
Then the series ∞∑n=Nan and the integral ∫∞Nf(x)dx either both converges or both diverges.
Calculation:
Given series is ∞∑n=N2n2+1
Using Limit comparing test,
limx→∞2n2+11n2=limx→∞2x2n2+1limx→∞2n2+11n2=2
The integral test applies because ∫∞N1xdx
By p-series test, ∫∞N1xpdx converges if p>1 , and diverges if p≤1 .
It is of the form of ∫∞N1xpdx where p=43 .
This series converges since p>1 .
The limit is positive, ∞∑n=N2n2+1 also converges.
b.
To find: To find whether the given series converges or diverges.
b.

Answer to Problem 1QQ
The series converges.
Explanation of Solution
Given:
The given series ∞∑n=N2n−13n+1 .
Formula used:
a=f(n)where f is a continuous, positive, decreasing function of x for all x≥N (N is a positive integer).
Then the series ∞∑n=Nan and the integral ∫∞Nf(x)dx either both converges or both diverges.
Calculation:
Given series is ∞∑n=N2n−13n+1 .
Using Limit comparing test,
limx→∞2n−13n+12n3n=limx→∞2n−12n3n+13nlimx→∞2n−13n+12n3n=1
If the sequence of partial sum has a limit as n→∞ , series convergence otherwise the sequence is divergence.
S=∞∑n=N2n3n
This is a geometric series whose initial term is zero and common ratio is 12 .
S=∞∑n=N2n3n=231−23S=2
The limit is positive, ∞∑n=N2n2+1 also converges.
c.
To find whether the given series converges or diverges.
c.

Answer to Problem 1QQ
The series converges.
Explanation of Solution
Given:
The given series ∞∑n=N4√nn .
Formula used:
a=f(n)where f is a continuous, positive, decreasing function of x for all x≥N (N is a positive integer).
Then the series ∞∑n=Nan and the integral ∫∞Nf(x)dx either both converges or both diverges.
Calculation:
Given series is ∞∑n=N4√nn .
∞∑n=N4√nn
By p-series test, ∫∞N1xpdx converges if p>1 , and diverges if p≤1 .
It is of the form of ∫∞N1xpdx where p=34 .
This series converges since p=1 .
The limit is positive, ∫∞N4√nndx also divergence.
Chapter 9 Solutions
Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020
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