Difference between revisions of "Mathematical Methods in the Physical Sciences/Chapter 1/Section 5"
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&= \lim_{n \to \infty}(-1)^n \lim_{n \to \infty}{n^2 \over n^2-1} \\ | &= \lim_{n \to \infty}(-1)^n \lim_{n \to \infty}{n^2 \over n^2-1} \\ | ||
&= \lim_{n \to \infty}(-1)^n \lim_{n \to \infty}{2n \over 2n} \text{ (L'Hopital's Rule)} \\ | &= \lim_{n \to \infty}(-1)^n \lim_{n \to \infty}{2n \over 2n} \text{ (L'Hopital's Rule)} \\ | ||
| − | &= \lim_{n \to \infty}(-1)^n | + | &= \lim_{n \to \infty}(-1)^n \\ |
| + | \end{align} | ||
</math> | </math> | ||
Revision as of 19:11, 18 April 2012
Contents
Overview
Introducing the preliminary test!
Notes
- The preliminary test of the convergence of a series says if <math>\lim_{n \to \infty}a_n \ne 0</math>, then the series diverges.
- The preliminary test say nothing about convergence.
Problems
#1
Each term is <math>{(-1)^n n^2 \over n^2-1}</math>. The limit is:
<math> \begin{align} &\lim_{n \to \infty}{(-1)^n n^2 \over n^2-1} \\ &= \lim_{n \to \infty}(-1)^n \lim_{n \to \infty}{n^2 \over n^2-1} \\ &= \lim_{n \to \infty}(-1)^n \lim_{n \to \infty}{2n \over 2n} \text{ (L'Hopital's Rule)} \\ &= \lim_{n \to \infty}(-1)^n \\ \end{align} </math>
Series diverges.
