Difference between revisions of "Mathematical Methods in the Physical Sciences/Chapter 1/Section 5"

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(#2)
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&= \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}} \over 1} \text{ (Hello, L'Hopital!)} \\
 
&= \lim_{n \to \infty} {1 \over 2\sqrt{n+1}} \\
 
&= \lim_{n \to \infty} {1 \over 2\sqrt{n+1}} \\
 +
&= 0
 
\end{align}
 
\end{align}
 
</math>
 
</math>
 
The limit is 0 since the denominator will grow to infinity.
 
  
 
We don't know if it is convergent or not.
 
We don't know if it is convergent or not.

Revision as of 19:24, 18 April 2012

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

<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

<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}} \\ &= 0 \end{align} </math>

We don't know if it is convergent or not.

#3

#4

#5

#6

#7

#8

#9

#10

#11