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.