Education.com
Try
Brainzy
Try
Plus

# Solutions to Series Practice Problems for AP Calculus

(not rated)
By McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these solutions can be found at:

Series Practice Problems for AP Calculus

1. is a geometric series with an initial term of one and a ratio of . Since the ratio is less than one, the series converges, and .
2. By the ratio test, . , therefore converges.
3. Consider the integral . Integrate by parts, with u = x, du = dx, dv = exdx, and v = – ex. . Therefore, Since the improper integral converges, converges.
4. Use the limit comparison test, comparing to the series , which is known to diverge. Divide . The limit , and diverges, so the series diverges.
5. Use the ratio test, so the series converges absolutely.
6. Use the ratio test. , therefore converges.
7. Compare to the p-series with p = 2, . This series, , is term-by-term less than or equal to the p-series with p = 2. Since that p-series converges, converges.
8. The sum of the geometric series
9. For the alternating series approximated by s50, the maximum absolute error so .
10. Examine the ratio of successive terms. Since , the series will converge when |x| < 1 or –1 < x < 1. When x = 1, the series becomes . This series is term-by-term smaller than the p-series with p = 2; therefore the series converges. When x = – 1, the series becomes which also converges. Therefore, the interval of convergence is [–1, 1].
11. The ratio is and so the series will converge when |3x| < 1. This tells you –1 < 3x < 1 and . When , the series becomes , which is a convergent p-series. When , the series becomes , a convergent alternating series. Therefore the interval of convergence is .
12. Represent ln x by a Taylor series. Investigate the first few terms by finding and evaluating the derivatives and generating the first few terms. f(a) = ln a, , so ln x can be represented by the series Using the ratio test, . The series converges when < 1, that is, –1 Solving the inequality, you find –a < xa < a or 0 < x < 2a. When x = 0, the series becomes . Since diverges, this series diverges as well. When x = 2a, the series becomes . This alternating series converges; therefore, the interval of convergence is (0, 2a].
13. Calculate the derivatives and evaluate at
x = 1. f (x) = ex2 and f (1) = e.
f'(x) = 2xex2 and f' (1) = 2e.
f''(x) = 4x2ex2 + 2ex2 and f''(1) = 6e.
f'''(x) = 8x3ex2 + 12xex2 and
f(4)(1) = 76e. Then the function f(x) = e x2 can be approximated by . Simplifying .
14. . Find the
derivatives and evaluate at .
,
,
, and
. Then f (x) = cos x around can be approximated by or .
15. At x = e, f(e) = ln e = 1,
,
,
• , and
. f(x) = ln x can be approximated by .
16. Calculate the derivatives and evaluate at
,
,
• ,
,
. In general, f (n)(0) = n!, so the MacLaurin series . The series converges to .The series converges on (–1, 1). When x = 1, the series becomes which diverges. When x = – 1, the series become which diverges. Therefore the interval of convergence is (–1, 1).
17. x =5cos 3θ cos θ and y =5cos 3θ sin θ.
= – 5sin θ cos 3θ – 15cos θ sin 3θ
=5cos θ cos 3θ – 15sin θ sin 3θ.
18. .

### Ask a Question

150 Characters allowed

### Related Questions

#### Q:

See More Questions

### Today on Education.com

Top Worksheet Slideshows