1. 9.2 (Larson Book) Nth term test Geometric series Telescoping series Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2003

2. This is an example of an infinite series. 1 1 Start with a square one unit by one unit: This series converges (approaches a limiting value.) Many series do not converge:

3. In an infinite series: a 1, a 2,… are terms of the series. a n is the n th term. Partial sums: n th partial sum If S n has a limit as, then the series converges, otherwise it diverges.

4. The first requirement of convergence is that the terms must approach zero. n th term test for divergence diverges if fails to exist or is not zero. Note that this can prove that a series diverges, but can not prove that a series converges. Ex. 2: If then grows without bound. If then As, eventually is larger than, therefore the numerator grows faster than the denominator. The series diverges. (except when x=0) Note This test can only determine divergence not convergence

5. Using the nth term test Determine if the series diverges ∞ Σ (5/4) n n=1

6. Using the n th term test Determine if the series diverges ∞ Σ (5/4) n n=1 The series diverges by the n th term test Warning: the nth term test can not guarantee convergence only divergence

7. Geometric Series: In a geometric series, each term is found by multiplying the preceding term by the same number, r. This converges to if, and diverges if. is the interval of convergence. (the proof is on the next slide)

8. Proof of geometric sum Start with a geometric sequence S = a + ar + ar 2 + ar 3 + ar 4 +… Multiply both sides by r Sr = ar + ar 2 + ar 3 + ar 4 +… Subract the two series S-Sr = a S(1- r) = a S = a/(1- r) Note: this series converges for │r│< 1

9. Example 1: a r

10. a r Example 2:

11. The partial sum of a geometric series is: If then If and we let, then: The more terms we use, the better our approximation (over the interval of convergence.)

12. Another series for which it is easy to find the sum is the telescoping series. Ex. 6: Using partial fractions: Telescoping Series converges to 

13. Determine if the following series converge, diverge or can’t tell. State the test that you used ∞∞ ∞ Σ(n!/2 n )Σ ln ((n+1)/n)Σ 6 (3/4) n n=1n=1n=1

14. Determine if the following series converge, diverge or can’t tell. State the test that you used ∞∞ ∞ Σ(n!/2 n )Σ ln ((n+1)/n)Σ 6 (3/4) n n=1n=1n=1 divergesconvergesconverges Nth term testtelescopingGeometric (│r│< 1)

15. Find the values of x for which the series converges ∞ Σ (x 2 /(x 2 +4)) n n=1

16. Find the values of x for which the series converges ∞ Σ (x 2 /(x 2 +4)) n n=1 Converges for all values of x because the denominator will always be larger than the numerator Verify with Algebra This series is geometric with r = x 2 /(x 2 +4) -1< x 2 /(x 2 +4) < 1 solve separately -1< x 2 /(x 2 +4) x 2 /(x 2 +4) < 1 0< x 2 /(x 2 +4) + 1 x 2 /(x 2 +4) – 1< 0 0< 2x 2 +4 /(x 2 +4) -4/(x 2 +4) <0 All values of x All values of x

17. Find the values of x for which the series converges ∞ Σ (-1) n x 2n n=1

18. Find the values of x for which the series converges ∞ Σ (-1) n x 2n n=1 Geometric series converges when │r│< 0 -1 <x< 1

19. Homework: p. 614 1- 89 every other odd An infinite crowd of mathematicians enters a bar. The first one orders a pint, the second one a half pint, the third one a quarter pint... "I understand", says the bartender - and pours two pints. Hint for joke: recall from today’s lesson

20. Geometric series have a constant ratio between terms. Other series have ratios that are not constant. If the absolute value of the limit of the ratio between consecutive terms is less than one, then the series will converge. For, if then: if the series converges.if the series diverges.if the series may or may not converge.