Wednesday, May 9, 2012

Calculus Revisited #17: Sequences

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


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