Answers and comments for Math 242 Homework 10.


#1. Final answer is 1.

Integration by parts should work.


#2. Final answer is

tan(x) + (2/3)tan(x)^3 + (1/5)tan(x)^5


#3. Final answer is: Diverges to infinity.

There may be various ways to show this.
One possibility is to compare to another
integral. But probably easiest is to do
the substitution u = x^2 + 2x.


#4. Final answer is: Converges absolutely.

Best way is Basic Comparison:

|sin(2n)| < 1

|sin(2n)|/(1+2^n) < 1/(1+2^n) < 1/2^n


#5. Final answer is: Converges conditionally.

(a) Since ln(n)/sqrt(n) decreases to 0,
the series converges by the Alternating test.

(b) If we consider Sum(ln(n)/sqrt(n))
we can e.g. do basic comparison

ln(n)/sqrt(n) > 1/sqrt(n)

Sum(1/sqrt(n)) = Sum(1/n^{1/2}) is divergent.

(a) & (b) --> original series converges conditionally.


#6. Final answer: Interval is [-1,1]

When applying ratio test, we get L = |x|

Endpoint x = -1 gives convergent alternating series

Endpoint x = +1 gives convergent p-series


#7. Final answer: Interval is (2,4]

When applying ratio test, we get L = |x-3|

Series converges if |x-3| < 1
Series converges if -1 < x-3 < 1
Series converges if 2 < x < 4

Endpoint x = 4 gives convergent alternating series

Endpoint x = 2 gives divergent series (similar to harmonic)


#8(i).

x + x^2 + x^3 + x^4 + ...

or: sum of x^n (n goes from 1 to infinity)


#8(ii).

1 - (3x)^2/2! + (3x)^4/4! - (3x)^6/6! + ...

or: sum of (-1)^n (3x)^(2n)/(2n)! (n goes from 0 to infinity)


#8(iii).

x + x^2/1! + x^3/2! + x^4/3! + ...

or: sum of x^(n+1)/n! (n goes from 0 to infinity)