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## Sequence

A sequence is an ordered list of numbers (e.g., $$a_n$$); the numbers are called "elements" or "terms". Every convergent sequence is bounded, thus an unbounded sequence is divergent.

Sequence Test Converge Notes
Squeeze Theorem $$\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} c_n = L$$ then $$\lim\limits_{n \to \infty} b_n = L$$ $$a_n \le b_n \le c_n$$
Def 1, pg 692 $$\lim\limits_{n \to \infty} a_n = L$$
l'Hospital's Rule $$\lim\limits_{n \to \infty} \frac{f(x)}{g(x)} \Rightarrow \lim\limits_{n \to \infty} \frac{f'(x)}{g'(x)}$$ where $$f(x)$$ = numerator and $$g(x)$$ = denominator
Theorem 3, pg 693 if $$f(n) = a_n$$ then $$\lim\limits_{n \to \infty} f(x)=L$$
Theorem 6, pg 694 $$\lim\limits_{n \to \infty} | a_n | =0$$ then $$a_n$$ converges
Theorem 9, pg 696 $$\lim\limits_{n \to \infty} r^n = \begin{cases} 0, & \text{if } -1 < r < 1 \\ 1, & \text{if } r = 1 \end{cases}$$ Divergent for all other values of $$r$$
Theorem 12, pg 698 Every bounded ($$m \le a_n \le M$$), monotonic sequence is convergent The bounds exists for $$n \ge 1$$, also see Theorem 10 and 11

## Series

A series is the sum of the terms of a sequence: $$\sum\limits_{n=1}^\infty a_n$$.

Series Test Converge Diverge Notes
Divergence N/A $$\lim\limits_{n\to\infty} a_n \ne 0$$ Doesn't show convergence and the converse is not true
Integral if $$\int\limits_1^\infty f(x) dx$$ converges if $$\int\limits_1^\infty f(x) dx$$ diverges $$f(x)$$ must be positive, decreasing, and continous, also $$f(n) = a_n \text{ for all } n$$
Root $$\lim\limits_{n\to\infty}\sqrt[n]{|a_n|} = L < 1$$ $$\lim\limits_{n\to\infty}\sqrt[n]{|a_n|} = L > 1 \text{ or } \infty$$ inconclusive if $$L = 1$$
Ratio $$\lim\limits_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right| = L < 1$$ $$\lim\limits_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right| = L > 1 \text{ or } \infty$$ inconclusive if $$L = 1$$
Direct Comparison $$0 \le a_n \le b_n \text{ for all } n$$ and $$\sum\limits_{n=1}^{\infty} b_n$$ converges $$0 \le b_n \le a_n \text{ for all } n$$ and $$\sum\limits_{n=1}^{\infty} b_n$$ diverges $$a_n,b_n > 0$$
Limit Comparison $$\lim\limits_{n\to\infty} \frac{a_n}{b_n} = L$$ and $$\sum\limits_{n=1}^{\infty} b_n$$ converges $$\lim\limits_{n\to\infty} \frac{a_n}{b_n} = L$$ and $$\sum\limits_{n=1}^{\infty} b_n$$ diverges $$a_n,b_n > 0$$ and L is a positive constant, if L is $$\infty$$ or 0, then pick a different $$b_n$$
Absolute $$\sum\limits_{n=1}^{\infty} | a_n | = 0$$   Definition of absolutely convergent, the sum is independent of the order in which the terms are summed
Conditional $$\sum\limits_{n=1}^{\infty} | a_n |$$ diverges but $$\sum\limits_{n=1}^{\infty} a_n$$ converges   The sum is dependent of the order in which the terms are summed

## Common Series

Series Test Formula Converge Diverge Notes
Alternating $$\sum\limits_{n=1}^\infty (-1)^{n-1} a_n$$ $$0 < a_{n+1} \le a_n \text{ for all } n$$ and $$\lim\limits_{n \to \infty} a_n = 0$$ N/A
Geometric $$\sum\limits_{n=1}^\infty ar^{n-1}$$ $$|r| < 1$$ and converges to $$\frac{a}{1-r}$$ $$|r| \ge 1$$ finite sum of the first n terms: $$= \frac{a(1-r^n)}{1-r}$$
P-Series $$\sum\limits_{n=1}^\infty \frac{1}{n^p}$$ $$p > 1$$ $$p \le 1$$ cannot calculate sum
Power $$\sum\limits_{n=0}^\infty c_n (x-a)^n$$ $$\begin{array}{l} i, \text{converge if } x=a \\ ii, \text{converge for all } x \\ iii, \text{converge if } |x-a|<R \end{array}$$   $$R$$ is the radius of convergence, you need to check the end points for convergence too. Typically use Ratio Test.
Taylor $$\sum\limits_{n=0}^\infty \frac{f^n (a)}{n!} (x-a)^n$$ $$|x-a|<R$$ $$|x-a|>R$$ Taylor series is centered about a. Same note as power series
Maclaurin $$\sum\limits_{n=0}^\infty \frac{f^n (0)}{n!} (x)^n$$ $$|x|<R$$ $$|x|>R$$ A Macluarin series is a Taylor series centered about 0. Same note as power series

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