8.2. Infinite Series http://www.ck12.org
Determine if∑∞k= (^1) k−^83 converges or diverges.
Solution
Using thenth-Term Test, limk→+∞k−^83. Since the limit is 0, we cannot make a conclusion about convergence or
divergence.
Rules for Convergent Series, Reindexing
Rules
As with sequences, there are rules for convergent infinite series that help make it easier to determine convergence.
Theorem (Rules for Convergent Series)
- Suppose∑∞k= 1 ukand∑∞n= 1 vkare convergent series with∑∞k= 1 uk=S 1 and∑∞k= 1 vk=S 2.
Then∑∞k= 1 (uk+vk)and∑∞k= 1 (uk−vk)are also convergent where
∑∞k= 1 (uk+vk) =∑∞k= 1 (uk)+∑∞k= 1 (vk) =S 1 +S 2 and∑∞k= 1 (uk−vk) =∑∞k= 1 (uk)−∑∞k= 1 (vk) =S 1 −S 2
(The sum or difference of convergent series is also convergent.) - Letc 6 =0 be a constant.
Suppose∑∞k= 1 ukconverges and∑∞k= 1 uk=SThen∑∞k= 1 cukalso converges where.
∞
k∑= 1 cui=c∞
k∑= 1 uk=cSIf∑∞k= 1 ukdiverges, then∑∞k= 1 cukalso diverges.
(Multiplying by a nonzero constant does not affect convergence or divergence.)
Example 10
Find the sum of∑∞k= 1 ( 3 k^2 − 1 + 8 k^1 − 1 ).
Solution
Using the Rules Theorem,∑∞k= 1 ( 3 k^2 − 1 + 8 k^1 − 1 )=∑∞k= 13 k^2 − 1 +∑∞k= 18 k^1 − 1.
∑∞k= 13 k^2 − 1 is a convergent geometric series witha=2 andr=^13. Its sum is 1 −^213 =^223 =3.
∑∞k= 18 k^1 − 1 is a convergent geometric series witha=2 andr=^18. Its sum is 1 −^218 =^178 =^87.
Then∑∞k= 1 ( 3 k^2 − 1 + 8 k^1 − 1 )=∑∞k= 13 k^2 − 1 +∑i∞= 18 k^1 − 1 = 3 +^87 =^297
Example 11
Find the sum of∑∞k= 12 ( 6 k^5 − 1 ).
Solution
By the rules for constant in infinite series,∑∞k= 12 ( 6 k^5 − 1 )= (^2) ∑∞k= 16 k^5 − 1. The series∑∞k= 16 k^5 − 1 is a geometric series
witha=5 andr=^16. Note that, by the Theorem on convergence of geometric series, this series converges to 6, that
is∑∞k= 16 k^5 − 1 = 1 −^516 =^556 =6.
Then∑∞k= 12 ( 6 k^5 − 1 )= 2 × 6 =12.
Adding or subtracting a finite number of terms from an infinite series does not affect convergence or divergence.