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Series Cheat Sheet

Arithmetic Series

$${s_n = n \frac{a_1 + a_n}{2}}$$

$${s_n = n \frac{2a_1 + d (n-1)}{2}}$$

$${a_n = a_1 + (n -1)d}$$

Geometric Series

$${\sum_{n=0}^{\infty} a (r^n) = \frac{a}{1 -r}, |r| < 1}$$

$${\sum_{n=1}^{\infty} a (r^{n-1}) = \frac{a}{1 -r}, |r| < 1}$$

Divergence Test

If \({\lim_{n \to \infty} a_n \ne 0}\) then \({\sum a_n}\) diverges

Comparison Test

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.

P-series Test

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

Integral Test

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.

Limit Comparison Test

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.

Alternating Series Test

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.

Ratio Test

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.

Root Test

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.

Absolute Convergence

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.

Telescoping Series

Series whose terms will cancel to a finite number of terms

Example:

$${\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+1}) }$$

$${\require{cancel}= (1 - \cancel{\frac{1}{2}}) + (\cancel{\frac{1}{2}} - \cancel{\frac{1}{3}}) + (\cancel{\frac{1}{3}} - \cancel{\frac{1}{4}}) + (\cancel{\frac{1}{4}} - \cancel{\frac{1}{5}}) + \dots = 1}$$

Harmonic Series

Divergent infinite series of the form \({\sum \frac{1}{an+b}}\)

Example:

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

Power Series

$${\sum_{n=0}^{\infty} C_n (x - a)^n}$$

radius of convergence: \({|x - a| < R}\)

interval of convergence: \({a - R < x < a + R}\) but check the endpoints for equality.

$${R = \lim\limits_{n \to \infty} |\frac{C_n}{C_{n + 1}}|}$$

Power Series - Geometric Series

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

Taylor Series

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

MacLaurin Series

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

Some common power series:

$${e^x = \sum_{n=0}^{\infty} \frac{(x)^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}}$$

$${cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(x)^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}}$$

$${sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(x)^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}}$$

$${cosh(x) = \sum_{n=0}^{\infty} \frac{(x)^{2n}}{(2n)!} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!}}$$

$${sinh(x) = \sum_{n=0}^{\infty} \frac{(x)^{2n+1}}{(2n+1)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}}$$

Useful facts (credit blackpenredpen):

For \({n \to \infty, p > 0, b > 1}\)

$${ln(n) < \sqrt{n} < n^p < b^n < n! < n^n}$$

Definition of e:

$${\lim\limits_{n \to \infty} (1 + \frac{a}{n})^{bn} = e^{ab}}$$

Don't use L'Hospitals Rule!:

$${\lim\limits_{\theta \to 0} \frac{sin(\theta)}{\theta} = 1}$$