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notes/education/math/MATH1220 (calc II)/Sequences.md
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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:
$$ \sum_{n = 1}^\inifty ar^{n-1} = \sum_{n=0}^\infty ar^n $$
Converges to $\dfrac{a}{1-r}$ if $|r| < 1$ or diverges if $|r| >= 1$.
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