The result of continuing to add the terms of a sequence indefinitely is called an infinite series. Although it appears to mean “adding infinitely many numbers,” it is precisely defined as the limit of the partial sums.
Definition of an Infinite Series
For a sequence {a_n}, let the sum through the nth term (the partial sum) be S_n = a_1 + a_2 + … + a_n.
Infinite series Σ_{n=1}^∞ a_n = lim_{n→∞} S_n
Infinite Geometric Series
An infinite series formed from a geometric sequence with first term a and common ratio r is called an infinite geometric series.
- When |r| < 1: It converges, and its sum is a/(1 − r)
- When |r| ≧ 1 (and a ≠ 0): It diverges
(1) 1 + 1/2 + 1/4 + 1/8 + … (2) 3 − 1 + 1/3 − 1/9 + …
Solution:
(1) a=1, r=1/2 → 1/(1−1/2) = 2
(2) a=3, r=−1/3 → 3/(1+1/3) = 9/4
Necessary Condition for a Convergent Series
If the series Σ a_n converges, then lim_{n→∞} a_n = 0. The converse is not true; the harmonic series Σ 1/n diverges.
Series Using Partial Fraction Decomposition
When 1/(n(n+1)) = 1/n − 1/(n+1), intermediate terms cancel each other (telescoping).
S_N = 1 − 1/(N+1) → 1 (N→∞)
Find the repeating decimal 0.3 (3 repeats indefinitely) as the sum of an infinite geometric series.
Solution:0.3 + 0.03 + 0.003 + …, a=0.3, r=0.1. Sum = 0.3/0.9 = 1/3
- Infinite series = the limit of partial sums
- Infinite geometric series: when |r|<1, sum = a/(1−r)
- If a series converges, lim a_n = 0 (the converse is not true)
- Use cancellation through partial fraction decomposition
Practice Problems
- Find the sum of the infinite geometric series 2 + 4/3 + 8/9 + ….
- When the infinite geometric series a/(1−r) has sum 6 and common ratio r=1/3, find a.
- Find Σ_{n=1}^∞ 1/((2n−1)(2n+1)).
Answers and Explanations
- Answer:6
Explanation: a=2, r=2/3. 2/(1−2/3) = 6. - Answer:4
Explanation: a/(2/3) = 6 → a = 4. - Answer:1/2
Explanation: (1/2){1/(2n−1) − 1/(2n+1)}. The partial sum (1/2)(1 − 1/(2N+1)) → 1/2.