• Sequence: Ordered list of numbers {an}
• Series: Sum of sequence terms: $\sum_{n=1}^\infty a_n$

Partial sums: $S_N = \sum_{n=1}^N a_n$

Series converges if $\lim_{N \to \infty} S_N$ exists (finite).

|r| < 1 for Convergence

Geometric series: $\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$ for |r| < 1

If lim (n -> ∞) an ≠ 0, then Sigmaan diverges.

Note: If limit = 0, test is inconclusive (series may or may not converge).

If f(n) = an, f positive, continuous, decreasing on [k, ∞): $\sum_{n=k}^\infty a_n \text{ and } \int_k^\infty f(x)dx$ both converge or both diverge.

p-series: Σ\SigmaΣ(1/n^p), converges if p > 1

$\sum_{n=1}^\infty \frac{1}{n^p}$

• Converges if p > 1
• Diverges if p ≤ 1

Harmonic series (p = 1): Diverges

If 0 ≤ an ≤ bn:

• $\sum b_n$ converges -> $\sum a_n$ converges
• $\sum a_n$ diverges -> $\sum b_n$ diverges

If $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ where 0 < c < ∞:

• Both series have same convergence behavior

For $\sum_{n=1}^\infty (-1)^{n-1}a_n$ or $\sum_{n=1}^\infty (-1)^n a_n$:

• an positive
• an decreasing
• $\lim_{n \to \infty} a_n = 0$

Then series converges.

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$

• L < 1: Converges absolutely
• L > 1: Diverges
• L = 1: Inconclusive

Limit L < 1, >1, or =1

Different outcomes based on L value.

If $\sum |a_n|$ converges, then $\sum a_n$ converges absolutely.

Conditional convergence: $\sum a_n$ converges but $\sum |a_n|$ diverges.

Addition, subtraction, multiplication, division (carefully), composition.

Taylor Polynomial Formula centered at x = a

$P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$

$|R_n(x)| \leq \frac{M|x-a|^{n+1}}{(n+1)!}$

where M = max|f(n+1)| on interval between x and a.

For $\sum_{n=0}^\infty c_n(x-a)^n$: $R = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right|$

Interval: (a - R, a + R); test endpoints separately.

Maclaurin (a = 0)

$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$

Taylor Series Formula centered at x = a

$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n, \quad |x| < 1$

$\frac{d}{dx}\sum_{n=0}^\infty c_n(x-a)^n = \sum_{n=0}^\infty \frac{d}{dx}[c_n(x-a)^n]$

$\int \sum_{n=0}^\infty c_n(x-a)^n dx = \sum_{n=0}^\infty \int c_n(x-a)^n dx$