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Limit Comparison Test: Determining Convergence of Series using Comparison Series, Lecture notes of Calculus

The limit comparison test (lct) is a method used to determine the convergence or divergence of a series by comparing it to a simpler series. The lct, its approach, and provides examples of its application.

Typology: Lecture notes

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Limit Comparison Test
Consider two series
โˆž
X
n=1
anand
โˆž
X
n=1
bnwith positive terms. Suppose that lim
nโ†’โˆž
an
bn
=C
with 0 < C < โˆž. Then
1. If
โˆž
X
n=1
bnconverges, then
โˆž
X
n=1
anconverges.
2. If
โˆž
X
n=1
bndiverges, then
โˆž
X
n=1
andiverges.
USED: When your given series behaves more like a simpler series, when nis large, but you may
not have a direct, obvious, and helpful bound as with the Comparison Test. Also only used for
positive termed series.
USED: LCT does not concern itself with which terms are bigger or smaller. LCT only cares if
the given terms and the comparison terms are about the same in size, as ngets very big. LCT is
the lazy version of comparison, and focuses on the idea that the original series and the (simpler?!)
comparison series share the same convergence behavior.
GOOD FOR: Xpolynomial
polynomial
NOTE: The order of the stack in the limit of an
bn
vs. bn
an
is not so important, because you are
just trying to decide if Cis finite and non-zero. It is often easier to put the given terms in the
numerator and the comparison terms in the denominator.
NOTE: If C= 0 or C=โˆž, then you either made a mistake in the Limit computation, OR chose
the wrong comparison series to start. Ignore all non-dominant terms as ngrows large, and try
again.
APPROACH:
โ€ขGiven the original series, start by ignoring non-dominant terms and decide what the compar-
ison series will be. Again, this comparison series is usually a p-series or a geometric series.
โ€ขCompare the terms. Compute the Limit of the stack of the given terms over the comparison
terms that are simpler. Justify the limit answer carefully. No guesses. If you use Lโ€™H Rule,
you must switch to the related function and the xvariable.
โ€ขAnalyze the comparison series completely.
โ€ขMake a conclusion about convergence for the original series.
pf2

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Limit Comparison Test

Consider two series

โˆ‘^ โˆž

n=

an and

โˆ‘^ โˆž

n=

bn with positive terms. Suppose that (^) nlimโ†’โˆž

an bn

= C

with 0 < C < โˆž. Then

  1. If

โˆ‘^ โˆž

n=

bn converges, then

โˆ‘^ โˆž

n=

an converges.

  1. If

โˆ‘^ โˆž

n=

bn diverges, then

โˆ‘^ โˆž

n=

an diverges.

USED: When your given series behaves more like a simpler series, when n is large, but you may not have a direct, obvious, and helpful bound as with the Comparison Test. Also only used for positive termed series.

USED: LCT does not concern itself with which terms are bigger or smaller. LCT only cares if the given terms and the comparison terms are about the same in size, as n gets very big. LCT is the lazy version of comparison, and focuses on the idea that the original series and the (simpler?!) comparison series share the same convergence behavior.

GOOD FOR:

โˆ‘ (^) polynomial polynomial

NOTE: The order of the stack in the limit of an bn vs. bn an is not so important, because you are

just trying to decide if C is finite and non-zero. It is often easier to put the given terms in the numerator and the comparison terms in the denominator.

NOTE: If C = 0 or C = โˆž, then you either made a mistake in the Limit computation, OR chose the wrong comparison series to start. Ignore all non-dominant terms as n grows large, and try again.

APPROACH:

  • Given the original series, start by ignoring non-dominant terms and decide what the compar- ison series will be. Again, this comparison series is usually a p-series or a geometric series.
  • Compare the terms. Compute the Limit of the stack of the given terms over the comparison terms that are simpler. Justify the limit answer carefully. No guesses. If you use Lโ€™H Rule, you must switch to the related function and the x variable.
  • Analyze the comparison series completely.
  • Make a conclusion about convergence for the original series.

EXAMPLES: Determine and state whether each of the following series converges or diverges. Name any convergence test(s) that you use, and justify all of your work.

โˆ‘^ โˆž

n=

n^3 + 7n n^9 +

n

Note that

โˆ‘^ โˆž

n=

n^3 + 7n n^9 +

n

โˆ‘^ โˆž

n=

n^3 n^9

โˆ‘^ โˆž

n=

n^6 which is a convergent p-series with p = 6 > 1.

Next, Check: lim nโ†’โˆž

n^3 + 7n n^9 +

n 1 n^6

= lim nโ†’โˆž

n^9 + 7n^7 n^9 +

n

n^9

n^9

) (^) = lim nโ†’โˆž

n^2 1 +

n

(^172)

= 1 which is finite and

non-zero.

Therefore, the O.S. is also Convergent, by Limit Comparison Test (LCT).

โˆ‘^ โˆž

n=

n^5 + n + 6 3 n^6 + n^4 + 5

โˆ‘^ โˆž

n=

n^5 n^6

โˆ‘^ โˆž

n=

n which is a divergent p-series with p = 1.

Next Check: lim nโ†’โˆž

n^5 + n + 6 3 n^6 + n^4 + 5 1 n

= lim nโ†’โˆž

n^6 + n^2 + 6n 3 n^6 + n^4 + 5

= lim nโ†’โˆž

1 + (^) n^14 + (^) n^65 3 + (^) n^12 + (^) n^56

which is finite and

non-zero

Therefore, these two series share the same behavior, and the Original Series is also divergent by Limit Comparison Test (LCT).

โˆ‘^ โˆž

n=

n n^6 + 2

โˆ‘^ โˆž

n=

n n^6

โˆ‘^ โˆž

n=

n^5

which is a convergent p-series with p = 5 > 1.

Next Check: (^) nlimโ†’โˆž

n n^6 + 2 1 n^5

= lim nโ†’โˆž

n^6 n^6 + 2 = lim nโ†’โˆž

1 + (^) n^26

= 1 which is Finite and Non-zero.

Therefore, O.S. is also convergent by LCT. (NOTE: CT would work too)

LAST NOTE: When the Limit C here is Finite and Non-Zero, the test does not directly conclude converge or diverge. LCT says if C is Finite and Non-Zero, then the given series has the same convergence or divergence behavior as its comparison series. The given series does whatever the comparison series does.