STRATEGY FOR TESTING SERIES
The main strategy is to classify the series according to its form.
1. If the series is of the form it is a p-series, which we know to be convergent if p > 1
1
np
∑,
and divergent otherwise.
2. If the series has the form or it is a geometric series, which converges if
arn−
∑1arn
∑,
and diverges otherwise. Some preliminary algebraic manipulation may be required to
| |r
<
1
bring the series into this form.
3. If the series has a form that is similar to a p-series or a geometric series, then one of the
comparison tests should be considered. In particular, if is a rational function, then the
an
series should be compared with a p-series. The value of p should be chosen by keeping
only the highest powers of n in the numerator and denominator. The comparison tests apply
only to series with positive terms, but if has some negative terms, then we can apply
an
∑
the Comparison Test to and test for absolute convergence.
| |an
∑
4. If you can see at a glance that then the Test for Divergence should be used.lim ,
nn
a
→∞
≠
0
5. If the series is of the form or then the Alternating Series Test is an
( )
−
−
∑11n n
b( ) ,
−
∑1nn
b
obvious possibility.
6. Series that involve factorials or other products (including a constant raised to the nth power)
are often conveniently tested using the Ratio Test. Bear in mind that
a
aas n
n
n
+
→ → ∞
11
for all p-series and therefore all rational or algebraic functions of n. Thus the Ratio Test
should not be used for such series.
7. If = f(n), where is easily evaluated, then the Integral Test is effective (assuming
anfxdx( )
1
∞
∫
all the hypotheses of this test are satisfied).
MIXED PRACTICE
Determine if each of the following series is convergent or divergent. If a series is a convergent
geometric series, find its sum.
1. 2. 3.
− + − + −
81
100
9
10 110
9. . .
n=
∞
∑
1
−
−
31
π
n
n=
∞
∑
1
2
3n
4. 5. 6. 7.
n=
∞
∑
1
1
2 3n
+
n=
∞
∑
1
1
1n
−
n=
∞
∑
1
1 2
1 3
+
+
n
nn
=
∞
∑
2
( )
ln
−
−
11n
n n
8. 9. 10. 11.
n=
∞
∑
1
( ) cos
−
1nn
π
n=
∞
∑
1
e n
n−!
n=
∞
∑
1
( )
( )
−
+
+ +
+
1 5
1 4
1 1
2 2
n n
n
nn=
∞
∑
1
n
nn
!
