/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Use the Direct Comparison Test t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 49

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} e^{-n^{2}} $$

Short Answer

Expert verified
The series \( \sum_{n=0}^{\infty} e^{-n^{2}} \) converges by the Direct Comparison Test.

Step by step solution

01

Understand the Direct Comparison Test

The Direct Comparison Test (or Comparison Test) is a method used to test for the convergence of an infinite series. It states that if 0 鈮� a鈧� 鈮� b鈧� for all n, and the series of b鈧� converges, then the series of a鈧� also converges. If the series of b鈧� diverges, the series of a鈧� does too. In this case, a suitable b鈧� to compare with the given series needs to be selected.
02

Choose an Appropriate Series for Comparison

For the series \( \sum_{n=0}^{\infty} e^{-n^{2}} \), a logical choice would be the series \( \sum_{n=0}^{\infty} e^{-n} \). The reason being \( e^{-n^{2}} \) is always less than or equal to \( e^{-n} \) for all \( n \geq 0 \) since when you square a nonnegative number, you only increase or keep it the same (for zero), and the exponential function is decreasing for negative inputs, so \( e^{-n^{2}} \leq e^{-n} \) holds.
03

Apply the Direct Comparison Test

It's known from the properties of geometric series that the series \( \sum_{n=0}^{\infty} e^{-n} \) converges because it's a geometric series with \( -1 < r = e^{-1}<1 \). Therefore, by the Direct Comparison Test, since the terms of the series \( \sum_{n=0}^{\infty} e^{-n^{2}} \) are nonnegative and less than or equal to the corresponding terms of a convergent series, the given series \( \sum_{n=0}^{\infty} e^{-n^{2}} \) also converges.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{3^{n}}{n^{3}} $$

State the definitions of convergent and divergent series.

Define a geometric series, state when it converges, and give the formula for the sum of a convergent geometric series.

In an experiment, three people toss a fair coin one at a time until one of them tosses a head. Determine, for each person, the probability that he or she tosses the first head. Verify that the sum of the three probabilities is 1 .

The random variable \(\boldsymbol{n}\) represents the number of units of a product sold per day in a store. The probability distribution of \(n\) is given by \(P(n) .\) Find the probability that two units are sold in a given day \([P(2)]\) and show that \(P(1)+P(2)+P(3)+\cdots=1\). $$ P(n)=\frac{1}{3}\left(\frac{2}{3}\right)^{n} $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.