7.5
Alternating Series (Special Series)
Theorem 89 (Alternating Series Test) Let 0 The alternating series
∞ X =1 (−1) and ∞ X =1 (−1)+1
converges if the following two condition are met. () lim
→∞= 0 () +1 ≤ for all
Example 192 Determine the convergence or divergence of
∞
X
=1
(−1) 1
Example 193 Determine the convergence or divergence of
∞ X =1 (−2)−1 Solution: For ≥ 1 1 2 ≤ + 1 2−1 2 ≤ + 1 ( + 1) 2−1 ≤ 2 + 1 2 ≤ 2−1
Hence, +1 = ( + 1) 2 ≤ 2−1= for all. Furthermore, by L’Hopital’s
Rule, lim →∞ 2−1 = lim→∞ 1 2−1(ln 2) = 0 This implies lim →∞ 2−1 = 0 105
Example 194 () The alternating series ∞ X =1 (−1)+1( + 1) = 2 1− 3 2+ 4 3− 5 4 + 6 5 − · · · Then +1 ≤ But lim →∞ + 1 = 16= 0 Moreover, lim →∞ (−1)+1 + 1 6= 0
Thue the series is diverges. () The alternating series
2 1 − 1 1+ 2 2− 1 2 + 2 3 − 1 3 + 2 4 − 1 4+· · · or = ½ 2 for is odd; 1 for is even. Then lim →∞ = 0
But is not nondecreasing. Moreover
2−1− 2 = 1 for = 1 2 3 · · · Thus ∞ X =1 (−1)+1 = ∞ X =1 1 =∞ Alternating Series Remainder:
Theorem 90 If P∞=1(−1) is converges to and +1 ≤ then
| − | = || ≤ +1
where =P∞= +1(−1)
Example 195 Approximate the sum of the following series by its first six term ∞ X =1 (−1)+1 µ 1 ! ¶ = 1 1!− 1 2! + 1 3!− 1 4! +· · ·
Solution: Clearly the alternating series is converges. The sum of the first six terms is 6 = 1 1 − 1 2 + 1 6 − 1 24 + 1 120 − 1 720 ≈ 063194 and by the Alternating Series Remainder, you have
| − 6| = |6| ≤ 7 =
1
5040 ≈ 00002 Therefore
063194− 00002 ≤ ≤ 063194 + 00002
Absolute and Condotion Convergence (General Series): For instance, the series
∞ X =1 sin 2 = sin 1 12 + sin 2 22 + sin 3 32 +· · ·
has both positive and negative terms, yet it is not an alternating series. But,
we know ¯ ¯ ¯ ¯sin 2 ¯ ¯ ¯ ¯ ≤ 12 for ≥ 1
By compareson test, the seriesP∞=1¯¯sin 2
¯
¯ is convergence. But the question still is ”Does the original series converges ?” Answer: YES!
Theorem 91 If the series P|| converges, the the series
P
also
con-verges.
Proof. Since 0 ≤ +|| ≤ 2 || for all by compareson test, we have
the series
∞
X
=1
+||
is converges with the converges series
∞
X
=1
||
Furthermore, since = (+||) − || and ∞ X =1 = ∞ X =1 (+||) − ∞ X =1 ||
Thus the series is converges.
Remark 7 The converse of above Theorem is not ture. For a counterexam-ple ∞ X =1 (−1)+1
Definition 41 (Absolute and conditional convergence) 1. P is absolutely convergent if P|| converges.
2. P is conditionally convergent if P converges but P || diverges. Example 196 () ∞ X =1 (−1)(+1)2 3 () ∞ X =1 (−1) ln ( + 1) () ∞ X =1 (−1)! 2 () ∞ X =1 (−1) √ 108
