STRATEGIES FOR SERIES
MATH 2419 - SPRING 2009 April 20, 2009
1. Try the nth term test for divergence If limn->ooan
i= a
then the series
L an diverges.
n=1
00
Note: this test cannot
be used to show convergence - if limn->ooan = a you must use some other test to determine convergence or divergence! 2. Determine if you have a special type of series
(a) Is the series geometric? Geometric series look like:
£;
00
(_2)3k-1 (-3)2k+1
The test to deal with this type of series is the Geometric Series Test (GST):
1.
First get series to start with index
L'
n=O
00
This is a very important step!
If your series does not start at n = correct. n. Once series is in the form A. If B. If
a
then your sum will not be
L ar
11=0
00
n
then apply the GST:
1~r
a < Irl <
Irl
1, series converges to the sum S = ?:: 1, series diverges
(b) Is the series telescoping? Telescoping series look like
1) L (1k + 1 - -k--9 4 4 +
00
k=1
There is no test for this type of series. Telescoping series always converge, because there exists a positive integer n such that all but a finite number of terms cancel out. Be sure to state the series is convergent because it is a telescoping series!
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Next, find the sum. You may be asked to do this by writing the expression for the nth partial sum, Sn, then the sum S = limn--->oo See Sn' 9(e) in the Exam 3 Review of how to do this (this is the preferred method). Another way to find the sum is to write out enough terms to show where the series collapses, identify which terms survive, and add them up. Note: sometimes you have to use partial fraction decomposition to get a series into telescoping form. See section 8.5 to review this technique. (c) Is the series a p-series?
00
1
p-series look like: ~ LnP These series converge if p > 1 and diverge if a < p ::; 1. To prove convergence or divergence of p-series, use the Integral Test:
1. n=1
To apply the Integral Test, consider an = fen). The function f(x) must be continuous, positive, and decreasing for x ?: 1 in order to apply this test. To check if f(x) is decreasing, find the first derivative and verify rex) < a for all x ?: 1. These conditions certainly hold for p-series. If the function f(x) is integrable and the above three conditions hold, the Integral Test can be applied. Evaluate the improper integral oo I(x) dx. The series converges if the integral converges and diverges if the integral diverges. For example, the series Test is applicable. 00 J2 xl~x dx.
11.
Jl
L -1n nn
00
1
is not a p-series but the Integral
n=2
Here you would evaluate the improper integral
3. If your series looks like one of the above types but is a slight deviation from their form, and the nth term test for divergence is inconclusive, then a comparison test should be used. To determine an appropriate series for comparison, look at the series you have and take the highest powers of n in the numerator and denominator and simplify. This series you use for comparison must be a known convergent or divergent series. Some examples where comparison 00 1 00 2 00 2n2 - 1 tests are used are ~ -2--' ~ -3--' and ~ 3 5 2 . These would Ln +1 L n-5 L n + n+1
n=1 n=1 n=l
2
be compared with
00
L
n=l
00
1 n2 (convergent p-series),
L3
n=l
00
1
n
(convergent geometric
series), and (a) Direct
1.
L :3 (convergent n
n=l
1
p-series) respectively.
Comparison
00
Test: Let 0 < an S bn for all n. then
If
L bn converges,
n=l
L an converges (i.e., if the
n=l
00
"larger" series
converges, the "smaller" series must also converge).
11.
If
L an
n=l
00
diverges, then
L bn diverges
n=l
00
(Le., if the "smaller" series
diverges, the "larger" series must also diverge). (b) Limit Comparison Test: This test is more "user friendly" than the Direct Comparison Test. The series for comparison is found the same way as for the previous test. Suppose that an > 0, bn > 0, and limn
00
-+
oo (~)
00
=
L where L is finite
and positive.
Then the two series
L an and L bn either
n=l n=l
both converge
or both diverge. 4. Is the series alternating? ~ (_1)n+1 series look like: L...t--n n=l makes the series alternate in sign. Alternating It is the (-1 )n+l portion which
(a) First check for absolute convergence (i.e. check to see if
L lanl
n=l
00
con-
verges). The most effective test to check for absolute convergence is the Ratio Test: L Let
L an be a series with nonzero terms.
n=l
00
Compute limn
-+
oo
I a~:l I.
00.
A. The series converges absolutely if the limit is < 1. B. The series diverges if the limit is > 1 or if the limit equals C. The test is inconclusive if the limit = 1. 3
ii. If this series converges then the given series converges absolutely and you are done. (b) If the series in the previous step diverges then you must apply the Alternating Series Test (AST) to check for conditional convergence: i. Let an > O. The alternating series
2) -ltan
n=l
00
and
2:.) _1)n+lan
n=l
00
converge if the followingtwo conditions are met: A. limn-+oo an = 0 B. an+l ::;an for all n ii. If the series converges at this step then it converges conditionally 111. If either of these conditions fail then the series is divergent. 5. IMPORTANT: be sure that when you state convergence or divergence of a series also state by which test you are drawing your conclusions!
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