notes/education/math/MATH1220 (calc II)/Sequences.md

A sequence is defined as an ordered list of numbers.

• Sequences are ordered, meaning two sequences that contain the same values but in a different order are not equal.
• Sequences can be infinite if a rule is defined, i.e \{1, 1, 1, 1, ...\}; a_i = 1

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

1. a_n + b_n \to L + M
2. C*a_n \to CL
3. a_n b_n \to LM
4. \frac{a_n}{b_n} \to \frac{L}{M} holds true where all values are defined
5. 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
6. 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:

\lim_{n \to \infty} \frac{n(n+1)}{2} = \infty

Given the above info, the limit is non-zero, so we know that the series diverges.

Geometric Series

A geometric series of the form:

\sum_{n = 1}^\infty ar^{n-1} = \sum_{n=0}^\infty ar^n

Converges to \dfrac{a}{1-r} if |r| < 1 or diverges if |r| >= 1.

Examples:

Determine if the series \sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) diverges or converges. If it converges, state where.

1. Rewrite 7^{-n} as (\frac{1}{7})^n to be closer to the form ar^n

= \sum_{n=1}^\infty 35*(\frac{1}{7})^n*2^{n-1}

2. Pull a 7 out of the bottom, making use of the fact that (\frac{1}{7})^n = \frac{1}{7}(\frac{1}{7})^{n-1}

= \sum_{n=1}^\infty \frac{35}{7}(\frac{1}{7})^{n-1}* 2^{n-1} = \sum_{n = 1}^\infty \frac{35}{7}(\frac{2}{7})^{n-1}

3. This is now of the form \sum_{n=1}^\infty ar^{n-1}, so:

\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1-\frac{2}{7}}

Divergence Test

If \lim_{n \to \infty} a_n \ne 0 then $\sum_{n=1}^\infty a_n$diverges.

The divergence test only tells us that if the limit does not equal zero, then the series diverges. If the limit is zero, it doesn't necessarily mean the series converges.

Eventually converging/diverging

Sometimes a series is not continually positive for the entire series, meaning most tests on series do not apply. To get around this, you can split the series into two or more parts. A finite negative number + infinity is still infinity, and a finite negative number + a finite number is still a finite number.