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

1 is...

n p

n=1

? convergent if p > 1 ? divergent if p 1

Geometric Series

ar n-1 is... ? convergent if r < 1

n=1

? divergent if r 1

If an has a form that is similar to one of the above, see whether you can use the

comparison test: Comparison Test

Example:

1

n=1 n 2 + n

(Warning! This only works if an and bn are always positive.)

(i) If an bn for all n, and bn is convergent, then an is convergent. (ii) If an bn for all n, and bn is divergent, then an is divergent.

Pick

bn

=

1 n2

(p-series)

an

=

1 n2 + n

1 n2

,

and

Example:

1

n=1 2n - 1

Pick bn

=

1 2n

(geometric)

lim an = lim 1 2n n bn n 2n - 1 1

Consider a series bn so that the ratio

an bn cancels the dominant terms in

the numerator and denominator of an , as in the example to the left. If you

know whether bn converges or not,

try using the limit comparison test.

1 converges, so by

n 2

n=1

(i),

1

converges.

n=1 n 2 + n

= lim 1 = 1 > 0 n 1 -1 2n

Limit Comparison Test

1 converges, so 2n

n=1

1 converges.

n=1 2n - 1

(Warning! This only works if an and bn are always positive.)

If lim an = c > 0 (and c is finite), then b n n

an and

bn either both

converge or both diverge.

Some series will "obviously" not converge--recognizing these can save you a lot of time and guesswork.

Test for Divergence

If

lim

n

a

n

0 , then

an

n=1

is divergent.

n2 -1

Example: n=1 n 2 + n

lim

n

an

= lim n2 -1 n n2 + n

1- 1 = lim n2

n 1 + 1

=1 0

n

n2 -1

so

is divergent.

n=1 n 2 + n

Testing for Convergence or Divergence of a Series (continued)

If an can be written as a function with a "nice" integral, the integral test may prove useful:

Integral Test

If f (n) = an for all n and f (x) is continuous,

positive, and decreasing on [1, ) , then:

If f (x)dx converges, then an converges.

1

n=1

If f (x)dx is divergent, then an is divergent.

1

n=1

Example: (-1)n-1

n 2

n=1

n3 +1

(i) 1 = n + 1 , so 1 = n + 1 + 1 > n + 1

bn

n2

bn +1

(n + 1)2

n+ 1 n2

=

1 bn

, so

1 bn+1

1 bn

, so

bn+1

bn

(ii)

lim

n

n2 n3 +1

=

lim

n

n

1 +1

n2

=0

So

(-1) n-1

n=1

n 2 n3 +1

is convergent.

Example:

1

n=1 n 2 + 1

f (x) = 1 is continuous, x2 +1

positive, and decreasing on [1, ) .

1

t

dx = lim

1 dx

1 x2 +1

t 1 x 2 + 1

] = limtan -1 x t = lim tan -1 t -

t

1 t

4

= - = , so 1 is

24 4

n=1 n 2 + 1

convergent.

Alternating Series Test

If (i) bn+1 bn for all n and (ii)

lim

n

bn

= 0 , then

(-1)n-1bn is

n=1

convergent.

The following 2 tests prove convergence, but also prove the stronger fact that an

converges (absolute convergence).

Ratio Test

If lim an+1 < 1 , then a n n

an is absolutely convergent.

If lim an+1 > 1 or lim an+1 = , then

a n n

a n n

an is divergent.

If lim an+1 = 1, use another test. a n

n

Example: e-nn! n=1

lim an+1 = lim e-n-1 (n + 1)!

a n n

n e -n n!

= e-1 lim n + 1 = , so n

e-nn! is divergent.

n=1

nn

Example:

31+3n

n=1

lim n nn = lim n

3 n

1+3n

n 31 n33

=

1 n

27

lim

n

31

n

=

So

n n

is divergent.

31+3n

n=1

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

Root Test

If

lim n

n

an

< 1, then

an is absolutely convergent.

If

lim n

n

a n

> 1 or lim an+1 a n

n

= , then

an is divergent.

If

lim n

n

an

= 1, use another test.

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