Series: comparison test

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Created this as a copy of Series: ratio test.

Name	Status	Author	Last Modified
Series: divergence test	Ready to use	Ben Brawn	03/10/2017 07:52
Series: ratio test	Ready to use	Ben Brawn	03/10/2017 07:50
Series: comparison test	Ready to use	Ben Brawn	03/10/2017 07:49
Monotonic sequence theorem	Ready to use	Ben Brawn	03/10/2017 07:48

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Question statement

This question is about the comparison test for series.

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			Name	Type	Generated Value

seed	integer	0
percent	integer	100
limit	rational	4/3
ccondiv	string	div
cseed	string	pd3
cexpression	string	$\displaystyle\simplify{{d}k^{
coeff	list	[ 5, 12, 4, 8, 10, 9 ]
a	number	5
b	number	12
d	number	4
f	number	8
g	number	10
start	number	9
pc1	string	$\displaystyle\simplify{{a}/({
pc2	string	$\displaystyle\simplify{{a}/({
pc3	string	$\displaystyle\simplify{{b}*co
pc4	string	$\displaystyle\simplify{{b}*si
pd1	string	$\displaystyle\simplify{{a}ln(
pd2	string	$\displaystyle\simplify{{a}/({
pd3	string	$\displaystyle\simplify{{d}k^{
gc1	string	$\displaystyle\simplify[fracti
gc2	string	$\displaystyle\simplify{{b}*co
gc3	string	$\displaystyle\simplify{{b}*si

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You have a series $\sum t_k$ with positive terms. Each term in the series is greater than or equal to the corresponding term in another series of positive terms which is actually known to converge.

You have a series $\sum t_k$ with positive terms. Each term in the series is greater than or equal to the corresponding term in another series of positive terms which is actually known to diverge.

You have a series $\sum t_k$ with positive terms. Each term in the series is less than or equal to the corresponding term in another series of positive terms which is actually known to converge.

You have a series $\sum t_k$ with positive terms. Each term in the series is less than or equal to the corresponding term in another series of positive terms which is actually known to diverge.

What does the comparison test tell us about the series $\sum t_k$ ?

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This series converges.

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The comparison test is as follows. Suppose that $0\ge a_n\ge b_n$ for sufficiently large $n$ .

• If $a_n$ diverges, then $b_n$ also diverges.

• If $b_n$ converges, then $a_n$ also converges.

Notice, if $a_n$ converges or if $b_n$ diverges the test doesn't say anything, in these cases some might say the test is 'inconclusive' or 'fails'.

The comparison test ensures us that our series converges (because a larger one does).

The comparison test ensures us that our series diverges (because a smaller one does).

The comparison test doesn't tell us anything about this situation.

The comparison test ensures us that our series converges (because a larger one does).

The comparison test ensures us that our series diverges (because a smaller one does).

The comparison test doesn't tell us anything about this situation.

In the denominator of {cexpression} the dominant term is $\var{d}k^\var{b}$ , so we will compare our series with $\sum_{k=\var{start}}^\infty\frac{\var{a}}{\var{d}k^\var{b}}$ which is a $p$ -series with $p=\var{b}$ and hence is convergent. Now for $k\ge\var{start}$ we definitely have

$\simplify{{a}/({d}k^{b}+{f}k+{g})<{a}/({d}k^{b})}$

and therefore our series is also convergent.

$\simplify{{a}/({d}k^{b}-{f}k-{g})>{a}/({d}k^{b})}$

and so can't use the comparison test to compare it to that series. But, for $k\ge\var{start}$ we definitely have

$\simplify{{a}/({d}k^{b}-{f}k-{g})<{a}/({d}k^{b+1})}$

and $\sum_{k=\var{start}}^\infty \simplify{{a}/({d}k^{b+1})}$ is a convergent $p$ -series. Therefore our series is also convergent.

Notice that $0\le \cos^2(k)\le 1$ and so for $k\ge\var{start}$ we definitely have

$\simplify{{b}cos(k)^2/({d}k^{a})<={b}/({d}k^{a})}$

Also notice, $\sum_{k=\var{start}}^\infty \frac{\var{b}}{\var{d}k^{\var{a}}}$ is a convergent $p$ -series with $p=\var{a}$ . Therefore our series is also convergent.

Notice that $0\le \sin^2(k)\le 1$ and so for $k\ge\var{start}$ we definitely have

$\simplify{{b}sin(k)^2/({d}k^{a})<={b}/({d}k^{a})}$

Also notice, $\sum_{k=\var{start}}^\infty \frac{\var{b}}{\var{d}k^{\var{a}}}$ is a convergent $p$ -series with $p=\var{a}$ . Therefore our series is also convergent.

For $k\ge\var{start}$ we definitely have

$\simplify{{a}ln(k)/({d}k)>{a}/({d}k)}$

and $\sum_{k=\var{start}}^\infty \simplify{{a}/({d}k)}$ is a divergent series (it is a scalar multiple of the harmonic series or a $p$ -series with $p=1$ ). Therefore our series is also divergent.

For $k\ge\var{start}$ we definitely have

$\simplify{{a}/({d}ln(k))>{a}/({d}k)}$

and $\sum_{k=\var{start}}^\infty \simplify{{a}/({d}k)}$ is a divergent series (it is a scalar multiple of the harmonic series or a $p$ -series with $p=1$ ). Therefore our series is also divergent.

Given $\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))}$ we might realise that

$\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))<{d}k^{b-1}/(sqrt(k^{2*b}))={d}k^{b-1}/(k^{b})={d}/k}$

however, $\sum \simplify{{d}/k}$ is divergent (it is a scalar multiple of the harmonic series or a $p$ -series with $p=1$ ) and given the direction of the inequality we can't use the comparison test to test these two series. But, lets try something else:

$\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))>{d}k^{b-1}/(sqrt(k^{2*b}+{a}k^{2*b}))={d}k^{b-1}/(sqrt({a+1}k^{2*b}))={d}k^{b-1}/(sqrt({a+1})k^{b})={d}/(sqrt({a+1})k)}$

Notice, $\sum_{k=\var{start}}^\infty \simplify{{d}/(sqrt({a+1})k)}$ is a divergent series (it is a scalar multiple of the harmonic series or a $p$ -series with $p=1$ ) and so our series is too.

In the denominator of {cexpression} the dominant term is $\var{d}^k$ , so we will compare our series with $\sum_{k=\var{start}}^\infty\frac{\var{a}}{\var{d}^k}$ which is the same as $\sum_{k=\var{start}}^\infty\var{a}\left(\frac{1}{\var{d}}\right)^k$ and so is a convergent geometric series with common ratio $r=\frac{1}{\var{d}}$ . Now for $k\ge\var{start}$ we definitely have

$\simplify{{a}/({d}^k+{f})<{a}/({d}^k)}$

and therefore our series is also convergent.

Notice that $0\le \cos^2(k)\le 1$ and so for $k\ge\var{start}$ we definitely have

$\simplify{{b}cos(k)^2/({d}{a}^k)<={b}/({d}{a}^k)}$

Also notice, $\sum_{k=\var{start}}^\infty \frac{\var{b}}{\var{d}\times\var{a}^k}$ is a convergent $p$ -series with with $p=\var{a}$ . Therefore our series is also convergent.

Notice that $0\le \sin^2(k)\le 1$ and so for $k\ge\var{start}$ we definitely have

$\simplify{{b}sin(k)^2/({d}*{a}^k)<={b}/({d}*{a}^k)}$

Also notice, $\sum_{k=\var{start}}^\infty \frac{\var{b}}{\var{d}\times\var{a}^k}$ is a convergent $p$ -series with with $p=\var{a}$ . Therefore our series is also convergent.

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This question is used in the following exams:

Series by Ben Brawn in Sequences and Series.
analysis by Daniel Atcheson in Daniel's workspace.
Series Homework 1 by Ugur Efem in Ugur's workspace.

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Ben Brawn 7 years, 7 months ago

Ben Brawn 7 years, 7 months ago

Ben Brawn 7 years, 7 months ago

Error in variable testing condition

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Series: comparison test

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Ben Brawn 7 years, 7 months ago

Ben Brawn 7 years, 7 months ago

Ben Brawn 7 years, 7 months ago

Error in variable testing condition

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Generated value: `integer`