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The concept of alternating series, including the definition, the alternating series test, and the estimation theorem. It also includes examples of various alternating series and their convergence. The document also discusses the Dirichlet test and its relation to the alternating series test.
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Learning Goals: Alternating Series
Alternating Series: Stewart Section 11.
Definition A series of the form
n=1(−1)
nbn or ∑∞ n=1(−1)
n+1bn, where bn > 0 for all n, is called an alternating series, because the terms alternate between positive and negative values.
We have already looked at an example of such a series in detail, namely the alternating harmonic series ∑^ ∞
n=
(−1)n−^1
n
. We proved that this series converges by showing that the even partial sums s 2 n form a
monotone bounded sequence and thus converge to a limit γ. We also showed that the odd partial sums S 2 n+1 must converge to the same limit γ and thus the series converges. The proof given applies to a more general class of alternating series which we will describe below. Further examples of alternating series are:
Example ∑∞
n=
(−1)n n
n=
(−1)n+^
n 2 n + 1
Note We can use the divergence test to show that the second series above diverges, since
lim n→∞
(−1)n+^
n 2 n + 1
does not exist
We have the following test for such alternating series:
Alternating Series test If the alternating series
∑^ ∞
n=
(−1)n−^1 bn = b 1 − b 2 + b 3 − b 4 +... bn > 0
satisfies (i) lim n→∞ bn = 0
(ii) bn+1 ≤ bn for all n then the series converges.
we see from the graph below that because the values of bn are decreasing, the partial sums of the series cluster about some point in the interval [0, b 1 ].
Click on the blue link to see a full proof similar to that given for the alternating harmonic series at the end of the notes.
Estimating the Error
Suppose
i=1(−1)
n− (^1) bn, bn > 0, converges to s. Recall that we can use the partial sum sn = b 1 −
b 2 + · · · + (−1)n−^1 bn to estimate the sum of the series, s. If the series satisfies the conditions for the Alternating series test, we have the following simple estimate of the size of the error in our approximation |Rn| = |s − sn|. (Rn here stands for the remainder when we subtract the n th partial sum from the sum of the series. )
Alternating Series Estimation Theorem If s =
(−1)n−^1 bn, bn > 0 is the sum of an alternating series that satisfies (i) bn+1 < bn for all n (ii) lim n→∞ bn = 0
then |Rn| = |s − sn| ≤ bn+1.
click on the blue link to see the proof included at the end of the notes.
Example Find a partial sum approximation the sum of the series
(−1)n^ n^1 where the error of approximation is less than .01 = 10−^2.
Example (a) Write down the Taylor series expansion of cos(x) about 0.
(b) Use part (a) to find the Taylor series expansion of cos(x^2 ) about 0.
(c) Use part (b) to express
0
cos(x^2 ) dx as a series.
(d) Use the alternating series estimation theorem to estimate
0
cos(x^2 ) dx with a maximum error of
10 −^8.
Click on the blue link to find solutions to a similar old exam question which uses the alternating series estimation theorem in conjunction with power series at the end of the lecture.
Proof of the Alternating Series Test
s 2 = b 1 − b 2 ≥ 0 since b 2 < b 1
s 4 = s 2 + (b 3 − b 4 ) ≥ s 2 since b 4 < b 3 .. . s 2 n = s 2 n− 2 + (b 2 n− 1 − b 2 n) ≥ s 2 n− 2
Hence the sequence of even partial sums is increasing:
s 2 ≤ s 4 ≤ s 6 ≤ · · · ≤ s 2 n ≤...
Also we have s 2 n = b 1 − (b 2 − b 3 ) − (b 4 − b 5 ) − · · · − (b 2 n− 2 − b 2 n− 1 ) − b 2 n ≤ b 1.
Hence the sequence of even partial sums is increasing and bounded and thus converges.. Therefore limn→∞ sn = s for some s. This takes care of the even partial sums, now we deal with the odd partial sums. We have s 2 n+1 = s 2 n + b 2 n+1, hence limn→∞ s 2 n+1 = limn→∞(s 2 n) + limn→∞ b 2 n+1) = limn→∞(s 2 n) = s, since by assumption (ii), limn→∞ b 2 n+1 = 0. Thus the limits of the entire sequence of partial sums is s and the series converges.
back to lecture
Note: that in the proof above we see that if s =
n=1(−1)
n− (^1) b n, with then
s 2 n ≤ s ≤ s 2 n+
because s 2 n+1 = s 2 n + b 2 n+1 and s = s 2 n + b 2 n+1 − (b 2 n+2 − b 2 n+3) − .... < s 2 n+1. Similarly in the proof above we see that s 2 n− 1 ≥ s ≥ s 2 n.
Proof of Alternating Series Estimation Theorem From our note above, we have that the sum of the series, s, lies between any two consecutive sums, and hence
|Rn| = |s − sn| ≤ |sn+1 − sn| = bn+1.
back to lecture
Old Exam Question
Part (a) Give the Taylor series expansion for the antiderivative
F (x) =
cos (
x) dx
about 0 (McLaurin Series) where F (0) = 0. Hint: Use your knowledge of a well known series. Solution to part (a): We know that the Taylor series expansion for cos(x) around x = 0 is cos(x) = ∑∞ n=
(−1)nx^2 n (2n)! , which has radius of convergence^ R^ =^ ∞.^ Plugging in^
x we obtain cos(
x) = ∑∞ n=
(−1)nxn (2n)! which is valid for all^ x^ ≥^ 0. Finally, we compute the indefinite integral
F (x) =
n=
(−1)nxn (2n)!
dx =
n=
(−1)n (2n)!
xndx
n=
(−1)n (2n)!
xn+ n + 1
Plugging in x = 0 we obtain
n=
(−1)n (2n)!
0 n+ n + 1
So, C = 0, and
F (x) =
n=
(−1)n (2n)!
xn+ n + 1
Part (b) Use part (a) to find an expression for the definite integral ∫ (^1)
0
cos(
x) dx
as a sum of an infinite series. Solution to part (b): By the Fundamental Theorem of Calculus we know ∫ (^1)
0
cos(
x) dx = F (1) − F (0)
n=
(−1)n (2n)!
1 n+ n + 1
n=
(−1)n (2n)!
n + 1
Part (c) Use the alternating series estimation theorem to estimate the value of the above definite integral so that the error of estimation is less than 1001. (you may write your answer as a sum of fractions). Solution to Part (c): The series in part (b) is of the form
n=0(−1)
nbn with bn = 1 (2n)!(n+1).^ We check that this series satisfies the conditions for the Alternating Series Estimation Theorem
