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.

If na has a form that is similar to one of the above, see whether you can use the comparison test:

Geometric Series

∑∞

=

−

1

1

n

nar is… • convergent if 1<r

• divergent if 1≥r

p-Series

∑∞

=1

1n

pn is… • convergent if 1>p

• divergent if 1≤p

Example: ∑∞

= +12

1n nn

Pick 2

1n

bn = (p-series)

22

11nnn

an ≤+

= , and

∑∞

=12

1n n

converges, so by

(i),∑∞

= +12

1n nn

converges.

Some series will “obviously” not converge—recognizing these can save you a lot of time and guesswork.

Test for Divergence

If 0lim ≠∞→ nn

a , then ∑∞

=1nna is divergent.

Example: ∑∞

= +−

12

2 1n nn

n

nnna

nnn +−

=∞→∞→ 2

2 1limlim 0111

11lim

2≠=

+

−=

∞→

n

nn

so ∑∞

= +−

12

2 1n nn

n is divergent.

Limit Comparison Test (Warning! This only works if na and nb are always positive.)

If 0lim >=∞→

cba

n

n

n (and c is finite), then ∑ na and ∑ nb either both

converge or both diverge.

Example: ∑∞

= −1 121

nn

Pick nnb21

= (geometric)

12

121limlim

n

nnn

n

n ba

−=

∞→∞→

01211

1lim >=−

=∞→ nn

∑∞

=1 21

nn converges, so

∑∞

= −1 121

nn converges.

Consider a series ∑ nb so that the ratio

nn ba cancels the dominant terms in the numerator and denominator of na , as in the example to the left. If you know whether ∑ nb converges or not, try using the limit comparison test.

Comparison Test (Warning! This only works if na and nb are always positive.) (i) If nn ba ≤ for all n, and ∑ nb is convergent, then ∑ na is convergent.

(ii) If nn ba ≥ for all n, and ∑ nb is divergent, then ∑ na is divergent.

If na can be written as a function with a “nice” integral, the integral test may prove useful:

Integral Test If nanf =)( for all n and )(xf is continuous, positive, and decreasing on [ )∞,1 , then:

If ∫∞

1

)( dxxf converges, then ∑∞

=1nna converges.

If ∫∞

1

)( dxxf is divergent, then ∑∞

=1nna is divergent.

Example: ∑∞

= +12 11

n n

11)( 2 +

=x

xf is continuous,

positive, and decreasing on [ )∞,1 .

dxx

dxx

t

t ∫∫ +=

+ ∞→

∞

12

12 1

1lim1

1

]4

tanlimtanlim 1

1

1 π−== −

∞→

−

∞→tx

t

t

t

442πππ

=−= , so ∑∞

= +12 11

n n is

convergent. Example: ∑∞

=

−

+−

13

21

1)1(

n

n

nn

(i) 2

11n

nbn

+= , so 1)1(

1112

1

+>+

++=+

nn

nbn

nbnn 11

2 =+≥ , so nn bb

11

1

≥+

, so nn bb ≤+1

(ii) 011lim

1lim 23

2

=+

=+ ∞→∞→ nnn

nnn

So ∑∞

=

−

+−

13

21

1)1(

n

n

nn is convergent.

Alternating Series Test If (i) nn bb ≤+1 for all n and (ii)

0lim =∞→ nn

b , then ∑∞

=

−−1

1)1(n

nn b is

convergent.

The following 2 tests prove convergence, but also prove the stronger fact that ∑ na converges (absolute convergence).

Ratio Test

If 1lim 1 <+

∞→n

n

n aa

, then ∑ na is absolutely convergent.

If 1lim 1 >+

∞→n

n

n aa

or ∞=+

∞→n

n

n aa 1lim , then ∑ na is divergent.

If 1lim 1 =+

∞→n

n

n aa

, use another test.

Root Test If 1lim <

∞→n

nna , then ∑ na is absolutely convergent.

If 1lim >∞→

nnn

a or ∞=+

∞→n

n

n aa 1lim , then ∑ na is divergent.

If 1lim =∞→

nnn

a , use another test.

Example: ∑∞

=

−

1!

n

nne

!)!1(limlim

11

nene

aa

n

n

nn

n

n −

−−

∞→

+

∞→

+=

∞=+=∞→

− 1lim1 nen

, so

∑∞

=

−

1!

n

nne is divergent.

Example: ∑∞

=+

1313n

n

nn

3131 33lim

3lim nn

nn

n

n

nn∞→+∞→

=

∞==∞→ nn

n13

lim271

So ∑∞

=+

1313n

n

nn is divergent.

When na contains factorials and/or powers of constants, as in the above example, the ratio test is often useful.

Testing for Convergence or Divergence of a Series (continued)