Comparison Test

Course:

fall-2010-math-2300-005

Section:

9.4

Date:

Tuesday, October 5, 2010 - 14:00 - Wednesday, October 6, 2010 - 15:00

Attachment	Size
fall2010math2300_comparison-test.pdf	46 KB

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Comparison Test Examples – October 5, 2010

Use the comparison test to determine if the given series converges or diverges.

$\sum _{{n=1}}^{\infty}\frac{1}{n^{4}+1}$

Solution: Since

$0\le\frac{1}{n^{4}+1}\le\frac{1}{n^{4}}$

and we know the $p$ -series

$\sum _{{n=1}}^{\infty}\frac{1}{n^{4}}$

converges, the given series also converges.
Use the comparison test to determine if the given series converges or diverges.

$\sum _{{n=1}}^{\infty}\frac{n-2}{n^{3}+2}$

Solution: We examine instead the convergence of

$\sum _{{n=2}}^{\infty}\frac{n-2}{n^{3}+2}$

(the same series, where the leading term has been removed). Since we know that removing a finite number of terms from a series will not affect convergence, this new series will converge if and only if the one in question converges. We notice that

$0\le\frac{n-2}{n^{3}+2}<\frac{n}{n^{3}+2}<\frac{n}{n^{3}}=\frac{1}{n^{2}}.$

(The reason we rewrite this series starting as 2 is because all of the terms need to be positive after a certain point in order to apply the comparison test… 2 is that point.) Since we know that the series

$\sum _{{n=2}}^{\infty}\frac{1}{n^{2}}$

converges, so does the series in question.
Use the comparison test to determine if the given series converges or diverges.

$\sum _{{n=1}}^{\infty}\frac{5}{2n^{2}+4n+3}$

Solution: Notice that

$\frac{5}{2n^{2}+4n+3}<\frac{5}{2n^{2}}$

and

$\sum _{{n=1}}^{\infty}\frac{5}{2n^{2}}=\frac{5}{2}\sum _{{n=1}}^{\infty}\frac{1}{n^{2}}$

where we know the series on the right converges. Thus, the series in question converges.
Use the comparison test to determine if the given series converges or diverges.

$\sum _{{n=1}}^{\infty}\frac{\ln n}{n}$

Solution: We used this example for the integral test as well. For the comparison test, we notice that when $n>e$ we have

$\frac{\ln n}{n}>\frac{1}{n}$

and we know that the $p$ -series

$\sum _{{n=1}}^{\infty}\frac{1}{n}$

diverges. Therefore, it follows from the comparison test that the given series diverges.

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