Series Convergence Notes

fall2010math2300_series-convergence-notes.pdf53.74 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Review on Series Convergence – October 6, 2010

§ Summary of Convergence of Series

§ Some General Notes

  • Pulling a finite number of terms off from a series will not affect convergence / divergence.

  • You can pull series apart like integrals (in terms of sums/difference and multiplication by a constant).

  • Series notation is shorthand, and in most cases series can be rewritten in any number of different ways.

    \sum _{{n=1}}^{\infty}\frac{(-1)^{n}}{n^{2}}=\sum _{{n=1}}^{\infty}\frac{(-1)^{{n+2}}}{n^{2}}=\sum _{{n=5}}^{\infty}\frac{(-1)^{{n-2}}}{(n-4)^{2}}=\sum _{{n=5}}^{\infty}\frac{(-1)^{n}}{(-1)^{2}(n-4)^{2}}=\sum _{{n=5}}^{\infty}\frac{\cos(n\pi)}{n^{2}-8n+16}

§ Limit of Series Terms   (Theorem 9.2 Part (c))

  • The series \displaystyle\sum _{{n=1}}^{\infty}a_{n} diverges if \displaystyle\lim _{{n\rightarrow\infty}}a_{n}\neq 0 or if \displaystyle\lim _{{n\rightarrow\infty}}a_{n} does not exist.

  • Conversely, just knowing \displaystyle\lim _{{n\rightarrow\infty}}a_{n}=0 doesn't tell us anything about convergence (except in the case of the Alternating Series Test, but there we also have to know that the magnitude of the terms is always getting smaller and the series is alternating).

§ Geometric Series

  • For finite geometric series \displaystyle\sum _{{n=1}}^{{k}}ax^{{n-1}}=a\frac{1-x^{k}}{1-x}.

  • For an infinite geometric series \displaystyle\sum _{{n=1}}^{\infty}ax^{{n-1}}=\frac{a}{1-x} provided that x<1.

§ p-series

  • The series \displaystyle\sum _{{n=1}}^{\infty}\frac{1}{n^{p}} converges if p>1 and diverges if p\le 1.

§ Integral Test

  • Let c be a positive integer and a_{n}=f(n) for positive integers n.

  • \displaystyle\sum _{{n=c}}^{\infty}a_{n} converges if \displaystyle\int _{c}^{\infty}f(x)\, dx converges.

  • \displaystyle\sum _{{n=c}}^{\infty}a_{n} diverges if \displaystyle\int _{c}^{\infty}f(x)\, dx diverges.

§ Comparison Test

  • Consider two series \displaystyle\sum _{{n=1}}^{\infty}a_{n} and \displaystyle\sum _{{n=1}}^{\infty}b_{n} with 0\le a_{n}\le b_{n}.

  • If \displaystyle\sum _{{n=1}}^{\infty}a_{n} diverges, then \displaystyle\sum _{{n=1}}^{\infty}b_{n} diverges.

  • If \displaystyle\sum _{{n=1}}^{\infty}b_{n} converges, then \displaystyle\sum _{{n=1}}^{\infty}a_{n} converges.

§ Limit Comparison Test

  • Consider two series \displaystyle\sum _{{n=1}}^{\infty}a_{n} and \displaystyle\sum _{{n=1}}^{\infty}b_{n} with a_{n}>0, b_{n}>0 and \displaystyle\lim _{{n\rightarrow\infty}}\frac{a_{n}}{b_{n}}=L where 0<L<\infty.

  • If one of the series diverges, then the other does as well.

  • If one of the series converges, then the other does as well.

§ Ratio Test

  • The series \displaystyle\sum _{{n=1}}^{\infty}a_{n} converges (absolutely) if \displaystyle L=\lim _{{n\rightarrow\infty}}\frac{|a_{{n+1}}|}{|a_{n}|}<1.

  • The series \displaystyle\sum _{{n=1}}^{\infty}a_{n} diverges if \displaystyle L=\lim _{{n\rightarrow\infty}}\frac{|a_{{n+1}}|}{|a_{n}|}>1.

  • The series \displaystyle\sum _{{n=1}}^{\infty}a_{n} may converge or may diverge if \displaystyle L=\lim _{{n\rightarrow\infty}}\frac{|a_{{n+1}}|}{|a_{n}|}=1.

§ Alternating Series Test

  • The alternating series \displaystyle\sum _{{n=1}}^{\infty}(-1)^{n}a_{n} converges if 0<a_{{n+1}}<a_{n} and \lim _{{n\rightarrow\infty}}a_{n}=0.

© 2011 Jason B. Hill. All Rights Reserved.