Calculus: Series and Sequences
Convergence, divergence, and power series
Calculus: Series and Sequences
Convergence, divergence, and power series
Calculus - Grade advanced
- 1
Determine whether the sequence a_n = (3n^2 - 5)/(2n^2 + n + 1) converges. If it converges, find its limit.
Compare the highest powers of n in the numerator and denominator.
The sequence converges to 3/2. Dividing the numerator and denominator by n^2 gives (3 - 5/n^2)/(2 + 1/n + 1/n^2), and as n approaches infinity, this expression approaches 3/2. - 2
Determine whether the sequence b_n = (-1)^n n/(n + 1) converges or diverges.
The sequence diverges. The factor n/(n + 1) approaches 1, so the terms behave like (-1)^n and oscillate between values near 1 and -1 rather than approaching one limit. - 3
Use the nth-term test to determine whether the series sum from n = 1 to infinity of n/(n + 2) diverges.
A convergent series must have terms that approach 0.
The series diverges by the nth-term test. The terms n/(n + 2) approach 1, not 0, so the infinite series cannot converge. - 4
Determine whether the geometric series sum from n = 0 to infinity of 5(2/3)^n converges. If it converges, find its sum.
The series converges because the common ratio is 2/3, and its absolute value is less than 1. The sum is a/(1 - r) = 5/(1 - 2/3) = 15. - 5
Determine whether the p-series sum from n = 1 to infinity of 1/n^(3/2) converges or diverges.
For a p-series, compare p to 1.
The series converges. It is a p-series with p = 3/2, and p-series converge when p is greater than 1. - 6
Use the integral test to determine whether the series sum from n = 2 to infinity of 1/(n ln n) converges or diverges.
Use the substitution u = ln x in the improper integral.
The series diverges. The integral of 1/(x ln x) from 2 to infinity is ln(ln x) evaluated from 2 to infinity, which grows without bound. - 7
Use the comparison test to determine whether the series sum from n = 1 to infinity of 1/(n^2 + 4) converges or diverges.
The series converges. Since 0 < 1/(n^2 + 4) < 1/n^2 for n at least 1, and the p-series sum 1/n^2 converges, the given series converges by comparison. - 8
Use the limit comparison test with sum from n = 1 to infinity of 1/n to determine whether the series sum from n = 1 to infinity of (3n + 1)/(n^2 + 5) converges or diverges.
The dominant behavior is approximately 3n/n^2 = 3/n.
The series diverges. The limit of [(3n + 1)/(n^2 + 5)] divided by 1/n is the limit of n(3n + 1)/(n^2 + 5), which equals 3. Since this positive finite limit compares the series to the divergent harmonic series, the given series diverges. - 9
Determine whether the alternating series sum from n = 1 to infinity of (-1)^(n + 1)/n converges absolutely, converges conditionally, or diverges.
The series converges conditionally. It converges by the alternating series test because 1/n decreases to 0, but it does not converge absolutely because the harmonic series sum 1/n diverges. - 10
Use the ratio test to determine whether the series sum from n = 1 to infinity of n!/5^n converges or diverges.
Compute the absolute value of a_(n+1)/a_n.
The series diverges. The ratio of consecutive terms is ((n + 1)!/5^(n + 1))/(n!/5^n) = (n + 1)/5, which approaches infinity. Since the ratio test limit is greater than 1, the series diverges. - 11
Use the root test to determine whether the series sum from n = 1 to infinity of (4n/(5n + 1))^n converges or diverges.
The series converges. The nth root of the absolute value of the nth term is 4n/(5n + 1), which approaches 4/5. Since 4/5 is less than 1, the series converges by the root test. - 12
Find the radius and interval of convergence for the power series sum from n = 0 to infinity of (x - 2)^n/3^n.
Rewrite the series as a geometric series and identify the ratio.
The series is geometric with ratio (x - 2)/3. It converges when |(x - 2)/3| < 1, so |x - 2| < 3. The radius of convergence is 3, and the interval of convergence is (-1, 5). At x = -1 and x = 5, the terms do not approach 0, so both endpoints are excluded. - 13
Find the radius and interval of convergence for the power series sum from n = 1 to infinity of n(x + 1)^n/4^n.
Using the ratio test, the limiting ratio is |x + 1|/4. The series converges when |x + 1| < 4, so the radius is 4. The possible interval is (-5, 3). At x = 3, the series becomes sum n, which diverges. At x = -5, the series becomes sum n(-1)^n, whose terms do not approach 0, so it diverges. The interval of convergence is (-5, 3). - 14
Write the first four nonzero terms of the Maclaurin series for e^x, then use it to approximate e^0.2.
The Maclaurin series for e^x is sum x^n/n! from n = 0 to infinity.
The first four nonzero terms are 1 + x + x^2/2! + x^3/3!. Substituting x = 0.2 gives 1 + 0.2 + 0.04/2 + 0.008/6 = 1.221333..., so e^0.2 is approximately 1.2213 using four nonzero terms. - 15
Find the Taylor series for f(x) = 1/(1 - x) centered at x = 0, and state its interval of convergence.
The Taylor series is 1 + x + x^2 + x^3 + ... = sum from n = 0 to infinity of x^n. It is a geometric series with ratio x, so it converges when |x| < 1. The interval of convergence is (-1, 1).