Mathematical Methods in the Physical Sciences/Chapter 1/Section 5

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.

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