Testing for Convergence or Divergence
Testing for Convergence or Divergence of a Series
Many of the series you come across will fall into one of several basic types. Recognizing
these types will help you decide which tests or strategies will be most useful in finding
whether a series is convergent or divergent.
p-Series
¡Þ
1
is¡
¡Æ
p
n =1 n
Geometric Series
? convergent if p > 1
? divergent if p ¡Ü 1
¡Þ
¡Æ ar
is¡ ? convergent if r < 1
n ?1
n =1
? divergent if r ¡Ý 1
If a n has a form that is similar to one of the above, see whether you can use the
comparison test:
¡Þ
1
Example: ¡Æ 2
Comparison Test
n =1 n + n
(Warning! This only works if a n and bn are always positive.)
1
Pick bn = 2 (p-series)
n
(i) If a n ¡Ü bn for all n, and ¡Æ bn is convergent, then ¡Æ a n is convergent.
1
1
(ii) If a n ¡Ý bn for all n, and ¡Æ bn is divergent, then ¡Æ a n is divergent.
¡Ü 2 , and
an = 2
n +n n
¡Þ
1
converges, so by
¡Æ
2
n
=1
n
Consider a series ¡Æ bn so that the ratio
¡Þ
¡Þ
1
1
Example: ¡Æ n
(i), ¡Æ 2
converges.
a n bn cancels the dominant terms in
n =1 2 ? 1
n =1 n + n
the numerator and denominator of a n ,
1
Pick bn = n (geometric)
as in the example to the left. If you
2
n
know whether ¡Æ bn converges or not,
a
1 2
lim n = lim n
try using the limit comparison test.
n ¡ú¡Þ b
n ¡ú¡Þ 2 ? 1 1
n
1
= lim
=1> 0
n ¡ú¡Þ 1 ? 1 2 n
Limit Comparison Test
¡Þ
(Warning! This only works if a n and bn are always positive.)
1
converges,
so
¡Æ
n
a
n =1 2
If lim n = c > 0 (and c is finite), then ¡Æ a n and ¡Æ bn either both
¡Þ
n ¡ú¡Þ b
1
n
converges.
¡Æ
n
converge
or both diverge.
n =1 2 ? 1
Some series will ¡°obviously¡± not
converge¡ªrecognizing these can save
you a lot of time and guesswork.
Test for Divergence
If lim a n ¡Ù 0 , then
n ¡ú¡Þ
n =1
1
n ?1
n2 = 1 ¡Ù 0
= lim
lim a n = lim 2
n ¡ú¡Þ
n ¡ú¡Þ
n ¡ú¡Þ n + n
1
1+
n
2
¡Þ
n ?1
so ¡Æ 2
is divergent.
n =1 n + n
2
¡Þ
¡Æa
n2 ?1
¡Æ
2
n =1 n + n
¡Þ
Example:
n
is divergent.
1?
Testing for Convergence or Divergence of a Series (continued)
If a n can be written as a function with a ¡°nice¡±
integral, the integral test may prove useful:
¡Þ
Example:
¡Æn
n =1
f ( x) =
positive, and decreasing on [1, ¡Þ ) , then:
If ¡Ò f ( x)dx converges, then
1
¡Þ
n =1
If ¡Ò f ( x)dx is divergent, then
1
n
Example:
¡Æ (?1)
n ?1
converges.
¡Þ
= lim tan ?1 x
¡Æ an is divergent.
t ¡ú¡Þ
n =1
n
n =1
= lim tan ?1 t ?
1
¦Ð
=
¦Ð
¦Ð
t ¡ú¡Þ
¡Þ
, so
¡Æn
n =1
2
4
1
is
+1
Alternating Series Test
If (i) bn +1 ¡Ü bn for all n and (ii)
lim bn = 0 , then
n ¡ú¡Þ
1
n2
= lim
=0
3
n ¡ú¡Þ n + 1
n ¡ú¡Þ n + 1 n 2
¡Æ (?1) n?1
?
t
3
¡Þ
¡Æ (?1)
n ?1
n =1
bn is
convergent.
(ii) lim
¡Þ
¦Ð
]
2 4 4
convergent.
2
n +1
1
1
1
1
(i) = n + 2 , so
= n +1+
> n +1
bn +1
bn
(n + 1) 2
n
1
1
1
1
, so
¡Ý n+ 2 =
¡Ý , so bn +1 ¡Ü bn
bn
bn +1 bn
n
n =1
So
t
1
1
dx
2
¡Ò1 x 2 + 1 dx = lim
¡Ò
t ¡ú¡Þ
1 x +1
=
¡Þ
2
¡Þ
¡Þ
¡Æa
1
+1
1
is continuous,
x +1
positive, and decreasing on [1, ¡Þ ) .
Integral Test
If f (n) = a n for all n and f (x) is continuous,
¡Þ
2
n2
is convergent.
n3 + 1
The following 2 tests prove convergence, but also prove the stronger fact that
¡Æa
n
converges (absolute convergence).
Ratio Test
a
If lim n +1 < 1 , then
n ¡ú¡Þ a
n
¡Þ
¡Æa
Example:
n
a
a n +1
> 1 or lim n +1 = ¡Þ , then
n ¡ú¡Þ a
n ¡ú¡Þ a
n
n
¡Æa
n
¡Þ
nn
¡Æ
1+ 3 n
n =1 3
nn
n
lim 1+3n = lim 1 n 3
n ¡ú¡Þ
n
¡ú
¡Þ
3
3 3
n
n
1
lim 1 n = ¡Þ
27 n¡ú¡Þ 3
¡Þ
nn
So ¡Æ 1+3n is divergent.
n =1 3
=
n!
a n +1
e ? n ?1 (n + 1)!
lim
= lim
n ¡ú¡Þ a
n ¡ú¡Þ
e ? n n!
n
is divergent.
= e ?1 lim n + 1 = ¡Þ , so
n ¡ú¡Þ
¡Þ
a
If lim n +1 = 1 , use another test.
n ¡ú¡Þ a
n
Example:
?n
n =1
is absolutely convergent.
If lim
¡Æe
¡Æe
?n
n! is divergent.
n =1
When a n contains factorials and/or powers of constants,
as in the above example, the ratio test is often useful.
Root Test
If lim n a n < 1 , then
n ¡ú¡Þ
¡Æa
n
is absolutely convergent.
a n +1
= ¡Þ , then
n ¡ú¡Þ a
n
If lim n a n > 1 or lim
n ¡ú¡Þ
If lim n a n = 1 , use another test.
n ¡ú¡Þ
¡Æa
n
is divergent.
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