Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{e^{n}}{\sqrt{n !}}$$

Short Answer

Expert verified
The series \( \sum_{n=0}^{\infty} \frac{e^n}{\sqrt{n!}} \) converges.

Step by step solution

01

Define the series terms

Identify the general term of the series, which is given as: \[a_n = \frac{e^n}{\sqrt{n!}}\]
02

Apply the Ratio Test

The Ratio Test requires evaluating the limit: \[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]Thus, calculate: \[a_{n+1} = \frac{e^{n+1}}{\sqrt{(n+1)!}}\]
03

Calculate the ratio \( \frac{a_{n+1}}{a_n} \)

Substitute \(a_{n+1}\) and \(a_n\) into the ratio: \[ \frac{a_{n+1}}{a_n} = \frac{ \frac{e^{n+1}}{\sqrt{(n+1)!}} }{ \frac{e^n}{\sqrt{n!}} } = \frac{e^{n+1}}{e^n} \cdot \frac{\sqrt{n!}}{\sqrt{(n+1)!}} \]Simplify this expression.
04

Simplify the ratio

Simplify the expression: \[ \frac{a_{n+1}}{a_n} = e \cdot \frac{\sqrt{n!}}{\sqrt{(n+1)!}} \]This can be further simplified as: \[ \frac{a_{n+1}}{a_n} = e \cdot \sqrt{ \frac{n!}{(n+1)!} } = e \cdot \sqrt{ \frac{1}{n+1} } \ = \frac{e}{\sqrt{n+1}} \]
05

Take the limit \( n \to \infty \)

Evaluate the limit: \[ L = \lim_{n \to \infty} \frac{e}{\sqrt{n+1}} \ = 0 \]
06

Determine convergence

Since \(L = 0\) and \(L < 1\), by the Ratio Test, the series \( \sum_{n=0}^{\infty} \frac{e^n}{\sqrt{n!}} \) converges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence of Series
Understanding whether a series converges or diverges is a foundational topic in mathematical analysis. A series is considered to converge if the sum of its infinite terms approaches a finite number as more terms are added. If this sum does not approach a finite number, the series is said to diverge. For convergence of series, various tests are applied to ascertain this behavior, and one such test is the Ratio Test.
Validators like the Ratio Test help determine if the infinite sum stabilizes at a finite value, ensuring stable mathematical expressions and meaningful results.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. Mathematically, it is represented as \(\sum_{n=0}^{\infty} a_n\) where \(a_n\) denotes the general term. The concept of infinite series is instrumental in various fields, such as calculus, physics, and engineering, where it helps model and solve complex problems.
Imagine adding up an endless list of numbers: if you can get a number at the end, your series converges! Otherwise, it diverges. Understanding infinite series allows for modeling continuous processes and capturing phenomena that evolve indefinitely.
For example, let's consider the series \(\sum_{n=0}^{\infty} \frac{e^n}{\sqrt{n!}}\): exploring its convergence involves understanding how quickly the terms shrink or grow, leading us to the next concept.
Mathematical Analysis
Mathematical Analysis is the branch of mathematics dealing with limits and related theories, such as differentiation, integration, measure, infinite series, and analytic functions. It provides tools to rigorously investigate variables that change continuously, the properties of sequences and series, and function behaviors.
Understanding convergence within the realm of mathematical analysis is essential as it provides a structured approach to grappling with infinite processes. It is about breaking down complex continuous changes into manageable analytical pieces.
Consider how precise limit calculations form the core of the Ratio Test method: it's all about using the tools of math analysis to simplify and clarify the behavior of series as their terms extend towards infinity.
Ratio Test for Series
The Ratio Test is a powerful technique to determine the convergence or divergence of an infinite series. It involves examining the limit of the absolute value of the ratio of consecutive terms of the series.
Here's a step-by-step walk-through based on the given exercise:
  • We start with identifying the general term, \(a_n = \frac{e^n}{\sqrt{n!}}\).
  • Next, we calculate the ratio of consecutive terms, \(\frac{a_{n+1}}{a_n} \), and simplify.
  • For our series, this ratio simplifies to \(\frac{e}{\sqrt{n+1}}\).
  • To determine convergence, we take the limit as \(n\) approaches infinity: \(L = \lim_{n \to \infty} \frac{e}{\sqrt{n+1}} = 0\).
  • Since \(L = 0 \text{and} L < 1\), we conclude that per the Ratio Test, the series converges.
Using the Ratio Test provides a systematic approach to solving convergence problems, making it essential for students tackling infinite series.

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Most popular questions from this chapter

The velocity \(v\) of electrons from a high energy accelerator is very near the velocity \(c\) of light. Given the voltage \(V\) of the accelerator, we often want to calculate the ratio \(v / c .\) The relativistic formula for this calculation is (approximately, for \(V \gg 1\) ) $$\frac{v}{c}=\sqrt{1-\left(\frac{0.511}{V}\right)^{2}}, \quad V=\text { number of million volts. }$$ Use two terms of the binomial series (13.5) to find 1 \(-v / c\) in terms of \(V\). Use your result to find \(1-v / c\) for the following values of \(V\). Caution: \(V=\) the number of million volts. (a) \(\quad V=100\) million volts (b) \(\quad V=500\) million volts (c) \(\quad V=25,000\) million volts (d) \(\quad V=100\) gigavolts \(\left(100 \times 10^{9} \text { volts }=10^{5}\) million volts) \right.

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{1-e^{x^{3}}}{x^{3}}$$

Test the following series for convergence. $$\sum_{n=0}^{\infty} \frac{(-1)^{n} n}{1+n^{2}}$$

Show that \(\sum_{n=2}^{\infty} 1 / n^{3 / 2}\) is convergent. What is wrong with the following "proof" that it diverges? $$\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}+\cdots>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+\cdots$$ which is $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\dots=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)$$. since the harmonic series diverges, the original series diverges. Hint: Compare \(3 n\) and \(n \sqrt{n}\).

There are 9 one-digit numbers ( 1 to 9 ), 90 two-digit numbers ( 10 to 99 ). How many three-digit, four-digit, etc., numbers are there? The first 9 terms of the harmonic series \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9}\) are all greater than \(\frac{1}{10}\); similarly consider the next 90 terms, and so on. Thus prove the divergence of the harmonic series by comparison with the series \(\left[\frac{1}{10}+\frac{1}{10}+\cdots\left(9 \text { terms each }=\frac{1}{10}\right)\right]+\left[90 \text { terms each }=\frac{1}{100}\right]+\cdots\) $$ =\frac{9}{10}+\frac{90}{100}+\cdots=\frac{9}{10}+\frac{9}{10}+\cdots $$ The comparison test is really the basic test from which other tests are derived. It is probably the most useful test of all for the experienced mathematician but it is often hard to think of a satisfactory \(m\) series until you have had a good deal of experience with series. Consequently, you will probably not use it as often as the next three tests.

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