This Part 17 of the Calculus Revisited series. Today we are dealing with sequences.
Sequence:
A sequence of numbers is exactly what you think of, a list of numbers. An infinite sequence is a list of numbers that goes on forever. In calculus, it is understood that unless specified, a sequence is an infinite one.
{a_n} = a_1, a_2, a_3, .... , a_n; n is an integer
Convergent Sequence:
A sequence {a_n} converges when
lim a_n = L
n → ∞
when L < ∞.
If the above condition is not met, then the sequence is said to diverge.
Bounded Sequence:
A sequence {a_n} is bounded if there exists a C such that
-C ≤ a_N ≤ C (alternatively |a_n| ≤ C)
If the above condition is not true, then the sequence is said to be unbounded.
Problems
1. Write the first five terms of the sequence {a_n} = 1 / n^3. Does this sequence converge?
a_1 = 1 / 1^3 = 1
a_2 = 1 / 2^3 = 1/8
a_3 = 1 / 3^3 = 1/27
a_4 = 1 / 4^3 = 1/64
a_5 = 1 / 5^3 = 1/125
lim 1 / n^3 = 0 as n → ∞. Conclude that the sequence converges.
Note that the sequence is also bounded. | 1 / n^3 | ≤ 1
2. Write the first five terms of {a_n} = (-1)^(n-1) * n^2. Does this sequence converge.
a_1 = (-1)^(1-1) * 1^2 = 1
a_2 = (-1)^(2-1) * 2^2 = -4
a_3 = (-1)^(3-1) * 3^2 = 9
a_4 = (-1)^(4-1) * 4^2 = -16
a_5 = (-1)^(5-1) * 5^2 = 25
lim (-1)^(n-2) * n^2 as n → ∞:
(-1)^(n-2) is bounded between -1 and 1.
However, n^2 grows to infinity.
Hence (-1)^(n-2) * n^2 → ∞ as n → ∞, the sequence diverges.
Also the sequence is unbounded.
3. Find the first six terms of the sequence {a_n} defined by the recursion formula:
a_0 = 1; a_n = 3 * a_n-1 + 2
a_0 = 1
a_1 = 3 (0) + 2 = 2
a_2 = 3 (2) + 2 = 8
a_3 = 3 (8) + 2 = 26
a_4 = 3 (26) + 2 = 80
a_5 = 3 (80) + 2 = 242
a_6 = 3 (242) + 2 = 728
Coming up: Series!
Eddie
This blog is property of Edward Shore. © 2012
