$${s_n = n \frac{a_1 + a_n}{2}}$$
$${s_n = n \frac{2a_1 + d (n-1)}{2}}$$
$${a_n = a_1 + (n -1)d}$$
$${\sum_{n=0}^{\infty} a (r^n) = \frac{a}{1 -r}, |r| < 1}$$
$${\sum_{n=1}^{\infty} a (r^{n-1}) = \frac{a}{1 -r}, |r| < 1}$$
If \({\lim_{n \to \infty} a_n \ne 0}\) then \({\sum a_n}\) diverges
For \({\sum a_n}\) and \({\sum b_n}\) where \({a_n \le b_n}\) for all n.
and \({a_n > 0}\) and \({b_n > 0}\) then
If \({\sum b_n}\) is convergent then \({\sum a_n}\) is convergent.
If \({\sum a_n}\) is divergent then \({\sum b_n}\) is divergent.
For \({\sum_\limits{n=0}^{\infty} \frac{1}{n^p}}\)
If \({p \gt 1}\) then the series converges
If \({p \le 1}\) then the series diverges
For \({\sum a_n}\) if \({a_n}\) is continous, positive and decreasing then
If \({\int_{k}^{\infty}{a_n \, dx} }\) is convergent then \({\sum a_n}\) is convergent.
If \({\int_{k}^{\infty}{a_n \, dx} }\) is divergent then \({\sum a_n}\) is divergent.
For \({\sum a_n}\) and \({\sum b_n}\)
if \({L = \lim\limits_{n \to \infty} \frac{a_n}{b_n}}\) where \({0 < L < \infty}\)
then either both \({\sum a_n}\) and \({\sum b_n}\) are convergent or
both series are divergent.
For \({\sum a_n}\) where \({a_n = (-1)^n b_n, b_n > 0}\)
Then if \({L = \lim\limits_{n \to \infty} {b_n = 0}}\) and \({b_n}\) is decreasing
then \({\sum a_n}\) is convergent.
For \({\sum a_n}\) and \({L = \lim\limits_{n \to \infty} {|\frac{a_{n+1}}{a_n}}|}\)
If \({L < 1}\) then \({\sum a_n}\) is absolutely convergent (hence convergent).
If \({L > 1}\) then \({\sum a_n}\) is divergent.
If \({L = 1}\) then the test is inconclusive.
For \({\sum a_n}\) and \({L = \lim\limits_{n \to \infty} {\sqrt[n]{|a_n|}}}\)
If \({L < 1}\) then \({\sum a_n}\) is absolutely convergent (hence convergent).
If \({L > 1}\) then \({\sum a_n}\) is divergent.
If \({L = 1}\) then the test is inconclusive.
If \({\sum |a_n|}\) is convergent then \({\sum a_n}\) is absolutely convergent.
If a series is absolutely convergent then it is also convergent.
If \({\sum |a_n|}\) is divergent and \({\sum a_n}\) is convergent then the series is conditionally convergent.
Series whose terms will cancel to a finite number of terms
Example:
$${\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+1}) }$$
$${\require{cancel}= (1 - \cancel{\frac{1}{2}}) + (\cancel{\frac{1}{2}} - \cancel{\frac{1}{3}}) + (\cancel{\frac{1}{3}} - \cancel{\frac{1}{4}}) + (\cancel{\frac{1}{4}} - \cancel{\frac{1}{5}}) + \dots = 1}$$
Divergent infinite series of the form \({\sum \frac{1}{an+b}}\)
Example:
$${\sum_{n=1}^{\infty} \frac{1}{n}}$$
$${\sum_{n=0}^{\infty} C_n (x - a)^n}$$
radius of convergence: \({|x - a| < R}\)
interval of convergence: \({a - R < x < a + R}\) but check the endpoints for equality.
$${R = \lim\limits_{n \to \infty} |\frac{C_n}{C_{n + 1}}|}$$
$${\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x}, |x| < 1, R = 1}$$
$${\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n}$$
$${\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} (x)^n}$$
$${e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots}$$
$${\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots}$$
$${\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}$$
$${\cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots}$$
$${\sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots}$$
$${\tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots}$$
$${\tanh^{-1}(x) = \sum_{n=0}^{\infty} \frac{ x^{2n+1}}{2n+1} = x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \cdots}$$
$${\frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) = \sum_{n=0}^{\infty} \frac{ x^{2n+1}}{2n+1} = x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \cdots}$$
$${\ln(1-x) = - \sum_{n=1}^{\infty} \frac{x^n}{n} = - (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots)}$$
$${\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots}$$
$${\psi(z) = -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+z} \right) = -\gamma + \sum_{n=0}^{\infty} \frac{z-1}{(n+1)(n+z)} }$$
$${\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+z} \right) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)} }$$
$${(1 + x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n}$$
$${(a + x)^k = \sum_{n=0}^{\infty} \binom{k}{n} (ax)^n}$$
$${\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots}$$
$${\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s} = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots}$$
$${\beta(s) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^s} = \frac{1}{1^s} - \frac{1}{3^s} + \frac{1}{5^s} - \frac{1}{7^s} + \cdots}$$
$${\eta(1) = \ln(2)}$$
$${\beta(1) = \frac{\pi}{4}}$$
$${\zeta(2) = \frac{\pi^2}{6}}$$
$${\text{even terms of } \zeta(2) = \frac{\pi^2}{24}}$$
$${\text{odd terms of } \zeta(2) = \frac{\pi^2}{8}}$$
$${\eta(2) = \frac{\pi^2}{12}}$$
$${\beta(2) \approx 0.915966... \text{(Catalan's constant)}}$$
$${\zeta(3) \approx 1.202057... \text{(Apery's constant)}}$$
For \({n \to \infty, p > 0, b > 1}\)
$${\ln(n) < \sqrt{n} < n^p < b^n < n! < n^n}$$
$${\lim\limits_{n \to \infty} (1 + \frac{a}{n})^{bn} = e^{ab}}$$