http://www.ck12.org Chapter 8. Infinite Series
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Intuitively, we can see that there is no bound on the sequence of partial sums and so the series diverges. This is
confirmed by the fact that the ratio of the series,r=^32 , tells us that the geometric series does not converge.
Another example of an important series is thep−series:
∑∞k= (^1) K^1 p=^1 + 21 p+ 31 p+ 41 p+, wherep>0.
Thep−series may look like a harmonic series, but it will converge for certain values ofp.
Theorem
Thep−series∑∞k= (^1) K^1 pconverges forp>1 and diverges for 0<p≤1.
Example 4Determine if∑∞k= 1 √^1 Kconverges or diverges.
Solution
Rewrite √^1 K asK^112 to get∑k∞= 1 √^1 K=∑+k=∞ (^1) K^112. The value ofpis^12. This is less than 1, which tells us that the
series diverges.
Comparison Test
Now that we have studied series without negative terms, we can apply convergence tests made for such series. The
first test we will consider is theComparison Test. The name of the test tells us that we will compare series to
determine convergence or divergence.
Theorem (The Comparison Test)
Let∑∞k= 1 ukand∑∞k= 1 vkbe series without negative terms. Suppose thatu 1 ≤v 1 ,u 2 ≤v 2 ,...,ui≤vi,....
- If∑∞k= 1 vkconverges, then∑∞k= 1 ukconverges.
- If∑∞k= 1 ukdiverges, then∑∞k= 1 vkdiverges.
In order to use this test, we must check that for each indexk, everyukis less than or equal tovk. This is the
comparison part of the test. If the series with the greater terms,∑∞k= 1 vk, converges, than the series with the lesser
terms ,∑∞k= 1 uk, converges. If the lesser series diverges, then the greater series will diverge. You can only use the
test in the orders given for convergence or divergence. You cannot use this test to say, for example, that if the greater
series diverges, than the lesser series also diverges.
Example 5Determine whether∑∞k= (^1) K (^31) + 3 converges or diverges.
Solution
∑∞k= (^1) K (^31) + 3 looks similar to∑∞k= (^1) K^13 , so we will try to apply the Comparison Test. Begin by comparing each term.
For eachk,K (^31) + 3 is less than or equal toK^13 , so∑∞k= (^1) K (^31) + 3 ≤∑∞k= (^1) K^13. Since∑∞k= (^1) K^13 is a convergentp−series, then,
by the Comparison Test,∑∞k= (^1) K (^31) + 3 also converges.
Example 6Determine whether∑∞k= 1 √ (^41) k− 5 converges or diverges.
Solution
The series∑∞k= 1 √ (^41) k− 5 is similar to∑∞k= 1 √ (^41) k. Using the Comparison Test, √ (^41) k− 5 ≥ √ (^41) kfor allk. The series
∑∞k= 1 √ (^41) kdiverges since it is ap−series withp=^14. By the Comparison Test,∑∞k= 1 √ (^41) k− 5 also diverges.