Behavior

A sequence is considered increasing if a_n is smaller than a_{n+1} for all n.
A sequence is considered decreasing if a_n is greater than or equal to a_{n+1} for all n.
Sequences exist that do not fall into either category, i.e, a_n = (-1)^n
If the terms of a sequence grow \{a_n\} get arbitrarily close to a single number L as n grows larger, this is noted by writing: \lim_{n\to\infty} a_n = L OR

a_n \to L \text{ as } n \to \infty

and say that a_n converges to L. If no L exists, we say \{a_n\} diverges.

Properties of Sequences

The below properties assume two sequences are defined, a_n \to L and b_n \to M

a_n + b_n \to L + M
C*a_n \to CL
a_n b_n \to LM
\frac{a_n}{b_n} \to \frac{L}{M} holds true where all values are defined
If L = M and a sequence c_n exists such that a_n \le c_n \le b_n for all n, then c_n \to L = M
If a_n and b_n both approach infinity at a similar rate, \frac{a_n}{b_n} will approach an arbitrary value. This value can be found by rewriting \frac{a_n}{b_n} in such a manner that the end behavior of the series is more easily identifiable

For example, given the series c_n = \frac{n}{2n+1}, both the numerator and the denominator approach infinity at a similar rate. However, when the numerator and denominator are both multiplied by \frac{1}{n}, it becomes \frac{1}{2+\frac{1}{n}}, an equivalent sequence that more clearly converges on 1/2.

Sum of an infinite sequence

If f(x) is a function and \{a_n\} is a sequence such that f(n) = a(n), then we say f(x) agrees with the sequence \{a_n\}
If f(x) agrees with \{a_n\} then if \lim_{x \to \infty}f(x) = L then \lim_{n \to \infty}a_n = L
Given the above knowledge, we can apply L'Hospital's rule to sequences that seem to approach \frac{\infty}{\infty}.

Remember, L'Hospital's rule states that:

If you have a limit of the indeterminate form \dfrac{0}{0}, the limit can be found by taking the derivative of the numerator, divided by the derivative of the denominator.
\lim_{x \to 2} \dfrac{x-2}{x^2-4} = \lim_{x \to 2} \dfrac{1}{2x}
L'Hospital's Rule can also be used when both the numerator and denominator approach some form of infinity.
\lim_{x \to \infty} \dfrac{x^2-2}{3x^2-4} = \lim_{x \to \infty} \dfrac{2x}{6x}
The above problem can be solved more easily without L'Hospital's rule, the leading coefficients are 1/3, so the limit as x approaches \infty is 1/3. L'Hospital's rule cannot be used in any other circumstance.

Series

Vocabulary: A series is another name for a sum of numbers.

The Limit Test

You can break a series into partial sums:

\sum_{n=1}^\infty a_n = a_1 + 1_2 + a_3 + ...

Given the above series, we can define the following:

S_1 = a_1 = \sum_{i=1}^\infty a_i
S_2 = a_1 + a_2 = \sum_{i=1}^2 a_i
S_n = a_1 + a_2 + ... = \sum_{i=1}^n a_i
If the limit of S_n as S_n approaches \infty converges to L, then we write: \sum_{n=1}^\infty a_n = L

and say that the sum converges to L.

Examples

Prove that \sum_{n = 1}^\infty \frac{1}{2^n} = 1

S_1 = \frac{1}{2}
S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}
S_3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}
S_n = \frac{2^n - 1}{2^n} So:

\sum_{n=1}^\infty \frac{1}{2^n} = \lim_{n \to \infty}S_n = \lim_{n \to \infty} (1 - \frac{1}{2^n}) = 1

Using the limit test, determine whether the series \sum_{n = 1}^\infty n converges or diverges

S_1 = 1
S_2 = 1 + 2 = 3
S_n = \frac{n(n+1)}{2} So:

3.8 KiB Raw Blame History