Difference between revisions of "Mathematical Methods in the Physical Sciences/Chapter 1/Section 5"
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=== #2 === | === #2 === | ||
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+ | Each term is <math>{\sqrt{n+1} \over n}</math>. | ||
+ | |||
+ | <math> | ||
+ | \begin{align} | ||
+ | \lim_{n \to \infty} {\sqrt{n+1} \over n} | ||
+ | &= \lim_{n \to \infty} {{1 \over 2\sqrt{n+1}} \over 1} \text{ (Hello, L'Hopital!)} \\ | ||
+ | &= \lim_{n \to \infty} {1 \over 2\sqrt{n+1}} \\ | ||
+ | \end{align} | ||
+ | </math> | ||
+ | |||
+ | The limit is 0 since the denominator will grow to infinity. | ||
+ | |||
+ | Not divergent, but we don't know if it is convergent. | ||
=== #3 === | === #3 === |
Revision as of 19:23, 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.
- For some reason, you'll get a lot of mileage out of L'Hopital's Rule.
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.
#2
Each term is <math>{\sqrt{n+1} \over n}</math>.
<math> \begin{align} \lim_{n \to \infty} {\sqrt{n+1} \over n} &= \lim_{n \to \infty} {{1 \over 2\sqrt{n+1}} \over 1} \text{ (Hello, L'Hopital!)} \\ &= \lim_{n \to \infty} {1 \over 2\sqrt{n+1}} \\ \end{align} </math>
The limit is 0 since the denominator will grow to infinity.
Not divergent, but we don't know if it is convergent.