Geometric Series
A geometric series is a series where each term is a constant multiple of the previous term. The general form of a geometric series is:
S = a + ar + ar2 + ar3 + ...
where:
- a is the first term,
- r is the common ratio (ratio between a term and its preceding term),
- S is the sum of the series.
Sum of an Infinite Geometric Series
The sum S of an infinite geometric series exists if |r| < 1. The sum can be found using the formula:
S = a / (1 - r)
Divergence and Convergence in Geometric Series
Understanding the convergence and divergence of a geometric series is pivotal in various mathematical and real-world applications. When |r| < 1, the terms in the series get smaller and approach zero, thus the series converges. Conversely, if |r| >= 1, the terms either get larger or oscillate, causing the series to diverge.
Example 1: Finding the Sum
Consider the series: 5 + 10 + 20 + 40 + ...
Here, a = 5 and r = 2. Since |r| > 1, the series does not have a sum (it diverges).
Example 2: Finding the Sum
Consider the series: 4 + 2 + 1 + 0.5 + ...
Here, a = 4 and r = 0.5. Since |r| < 1, we can find the sum:
S = 4 / (1 - 0.5) = 8
Convergence Tests
Convergence tests are methods used to determine whether an infinite series converges (has a finite sum) or diverges (does not have a finite sum). Here are a few key convergence tests:
1. The Ratio Test
The ratio test states that for a series ∑an, if there exists a limit L such that:
L = lim(n → ∞) |a(n+1) / an|
- If L < 1, the series converges.
- If L > 1, the series diverges.
- If L = 1, the test is inconclusive.
Example: Using the Ratio Test
Consider the series: ∑(1/n!) from n=1 to ∞.
L = lim(n → ∞) |1/(n+1)! / 1/n!| = lim(n → ∞) n!/ (n+1)! = lim(n → ∞) 1/(n+1) = 0
Since L < 1, the series converges.
2. The Comparison Test
The comparison test involves comparing the series ∑an with another series ∑bn.
- If 0 ≤ an ≤ bn for all n in some range, and ∑bn converges, then ∑an also converges.
- If 0 ≤ bn ≤ an for all n in some range, and ∑bn diverges, then ∑an also diverges.
Example: Using the Comparison Test
Consider the series: ∑(1/n2) from n=1 to ∞.
Comparing it with the series ∑(1/n) (which is known to diverge as it is a harmonic series), we see that since 1/n2 < 1/n and the larger series diverges, we cannot conclude that the smaller series also diverges. Further tests or knowledge (p-series) would be needed.
3. The Root Test
The root test states that for a series ∑an, if there exists a limit L such that:
L = lim(n → ∞) |an|(1/n)
- If L < 1, the series converges.
- If L > 1, the series diverges.
- If L = 1, the test is inconclusive.
Example: Using the Root Test
Consider the series: ∑2n from n=1 to ∞.
L = lim(n → ∞) |2n|(1/n) = lim(n → ∞) 2 = 2
Since L > 1, the series diverges.
4. The Integral Test
The integral test states that if f is a continuous, positive, decreasing function from 1 to ∞ and an = f(n), then the series ∑an and the integral ∫f(x)dx from 1 to ∞ both converge or both diverge.
Example: Using the Integral Test
Consider the series: ∑(1/n2) from n=1 to ∞.
The function f(x) = 1/x2 is continuous, positive, and decreasing from 1 to ∞. Evaluating the integral:
∫(1/x2)dx from 1 to ∞ = [−1/x] from 1 to ∞ = −1/∞ + 1 = 1
Since the integral converges, the series also converges.
5. The Alternating Series Test
The alternating series test states that the alternating series ∑(−1)(n+1)an or ∑(−1)nan converges if:
- lim(n → ∞) an = 0
- a(n+1) ≤ an for all n.
Example: Using the Alternating Series Test
Consider the series: ∑(−1)(n+1) / n from n=1 to ∞.
- lim(n → ∞) 1/n = 0
- 1/(n+1) ≤ 1/n for all n.
Since both conditions are satisfied, the series converges.
FAQ
The Root Test is employed to ascertain the convergence of a series by examining the nth root of its terms. Specifically, we compute lim(n → ∞) |an|(1/n). If this limit, referred to as the limit supremum or lim sup, is less than 1, the series converges. If it is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive, and we cannot determine the convergence or divergence of the series from the Root Test alone. The Root Test is particularly useful for series where the terms are raised to the power of n.
No, a series cannot converge if its terms do not approach zero. This is a fundamental principle in the study of series and sequences. If the limit of the terms an as n approaches infinity is not zero, i.e., lim(n → ∞) an ≠ 0, the series ∑an diverges. This principle is often used as a preliminary check for divergence in a series before applying other more complex tests for convergence.
The harmonic series is the series ∑(1/n) for n=1 to ∞. Despite the terms of the harmonic series approaching zero as n approaches infinity, the series itself diverges. This can be demonstrated using the Comparison Test by comparing it with another divergent series or by using the Integral Test. The harmonic series is a classic example used in mathematical discussions about convergence and divergence because it intuitively seems like it should converge due to the decreasing term size, yet it does not, highlighting the nuanced nature of infinite series.
Absolute convergence refers to the convergence of a series ∑an when the series ∑|an| also converges. In other words, if the series of the absolute values of the terms converges, the original series is said to converge absolutely. Conditional convergence, on the other hand, occurs when the original series ∑an converges, but the series ∑|an| does not. This distinction is vital because absolutely convergent series retain their convergence status even when terms are rearranged, while conditionally convergent series may not.
The comparison test is a method used to determine the convergence or divergence of a series by comparing it to another series whose convergence status is known. If we have a series ∑an and another series ∑bn, and 0 ≤ an ≤ bn for all n in some range, then if ∑bn converges, so does ∑an. Conversely, if ∑an diverges, so does ∑bn. Essentially, if the series you are comparing to is known to converge, and your series is always less than or equal to it, your series also converges. Similarly, if the known series diverges and your series is always greater than or equal to it, your series also diverges.
Practice Questions
The given series is ∑(3n / n! ) from n=1 to ∞. To determine whether it converges or diverges, we can use the ratio test, which involves finding the limit of the ratio of the (n+1)th term to the nth term as n approaches infinity. The ratio test states that if the limit is less than 1, the series converges, if it is greater than 1, it diverges, and if it is equal to 1, the test is inconclusive.
Calculating the limit, we have:
lim(n → ∞) |(3(n+1) / (n+1)!) / (3n / n!)| = lim(n → ∞) |3(n+1) * n! / (3n * (n+1)!)| = lim(n → ∞) |3 / (n+1)| = 0
Since the limit is less than 1, by the ratio test, the series converges.
The given series is ∑(−1)(n+1) / n from n=1 to ∞, which is an alternating series. To determine its convergence or divergence, we can use the alternating series test. The alternating series test states that the series converges if lim(n → ∞) an = 0 and a(n+1) ≤ an for all n.
Calculating the limit, we have:
lim(n → ∞) 1/n = 0
And since 1/(n+1) ≤ 1/n for all n, both conditions of the alternating series test are satisfied. Therefore, the series converges.
