Difference is denoted with d, such that d = a_{n+1} - a_n
Sum: S_n = \frac{2a_1 + (n-1)d}{2}n
n-th term: a_n = a_1 + (n-1)d
Relation: a_n = \frac{a_{n-1} + a_{n+1}}{2}
Quotient is denoted with q, such that q = \frac{a_{n+1}}{a_n}
Sum: S_n = a_1\frac{1 - q^n}{1 - q}
n-th term: a_n = a_1q^{n-1}
Relation: a_n^2 = a_{n-1}a_{n+1}
|a_n| < M
A sequence is convergent if it has a limit otherwise it is divergent
L - limit of a sequence
\epsilon -
then: |a_n - L| < \epsilon
a_n converges to L if all of its subsequences converge to L
let a_n = (-1)^n then a_{2n} = (-1)^{2n} and a_{2n-1} = (-1)^{2n-1} are subsequences of a_n. They have different limits therefore a_n has no limit.
a_n \leq b_n \leq c_n
if \lim a_n = \lim c_n = L then \lim b_n = L
\lim \limits_{n \to \infty} (1 + \frac{x}{n})^n = e^x
Iff \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0 or \lim_{x \to c} f(x) = \pm \lim_{x \to c} g(x) = \pm \infty then \lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}
implicit usage: