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Sequences and Series

Sun 11 October 2015

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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