5 Steps to a 5 AP Calculus BC 2019

Series 353

Informal Principle

Given a rational expression containing a polynomial innas a factor in the numerator or
denominator, often you may delete all but the highest power ofnwithout affecting the
convergence or divergence behavior of the series. For example,

∑∞

n= 1

4 n^3 −n+ 1

n^5 + 7 n^2 − 6

behaves like

∑∞

n= 1

4 n^3

n^5

= 4

∑∞

n= 1

1

n^2

When choosing a series for the Limit Comparison Test, it is helpful to apply the Informal
Principle.
Example 1

Determine whether the series 1+

1

5

+

1

9

+

1

13

+···converges or diverges.

The given series 1+

1

5

+

1

9

+

1

13

+···=

∑∞

n= 1

1

4 n− 3

. Choose

∑∞

n= 1

1

n

for comparison.

Since it has a similar structure, and we know it is ap-series withp=1, it diverges. The

nlim→∞

1 /(4n−3)

1 /n

=nlim→∞

n

4 n− 3

=

1

4

. The limit exists and is greater than zero; therefore, since

∑∞

n= 1

1

n

diverges, 1+

1

5

+

1

9

+

1

13

+···=

∑∞

n= 1

1

4 n− 3

also diverges.

Example 2

Determine whether the series
∑∞
n= 1

n− 2

n^3

converges or diverges.

Compare to the known convergent p-series

∑∞

n= 1

1

n^2

. The limit limn→∞
(n−2)/n^3
1 /n^2

=

nlim→∞

(n−2)(n^2 )

(n^3 )

=nlim→∞

n− 2

n

=1. Since 0<nlim→∞

(n−2)/n^3

1 /n^2

<∞and

∑∞

n= 1

1

n^2

converges,

∑∞

n= 1

n− 2

n^3

also converges.

Example 3

Determine whether the series

∑∞

n= 1

1

√

n^3 − 3 n

converges or diverges. Using the informal prin-

ciple, you have

∑∞

n= 1

1

√

n^3

or

∑∞

n= 1

1

n^3 /^2

. (Note that

∑∞

n= 1

1

n^3 /^2

is ap-serieswithp>1 and therefore

it converges.)

Apply the limit comparison test and obtain limn→∞

1

n^3 /^2

1

√

n^3 − 3 n

= nlim→∞

√

n^3 − 3 n

n^3 /^2

=

nlim→∞

√

n^3 − 3 n

n^3 /^2

=nlim→∞

√

1 −

3

n^2

=1. Since 0 < nlim→∞

1

n^3 /^2

1

√

n^3 − 3 n

< ∞, and

∑∞

n= 1

1

n^3 /^2

converges, the series

∑∞

n= 1

1

√

n^3 − 3 n

converges.