Question

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

Short answer

The series $\sum _{n=0}^{\mathrm{\infty }}\frac{e^n}{\sqrt{n!}}$ converges.

Step-by-step solution

Define the series terms

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

Apply the Ratio Test

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

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.

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: $$\frac{a_{n+1}}{a_n}=e\cdot \sqrt{\frac{n!}{(n+1)!}}=e\cdot \sqrt{\frac{1}{n+1}}=\frac{e}{\sqrt{n+1}}$$

Take the limit $n\to \mathrm{\infty }$

Evaluate the limit: $$L=\underset{n\to \mathrm{\infty }}{lim}\frac{e}{\sqrt{n+1}}=0$$

Determine convergence

Since $L=0$ and $L<1$, by the Ratio Test, the series $\sum _{n=0}^{\mathrm{\infty }}\frac{e^n}{\sqrt{n!}}$ converges.

Key concepts

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 $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ 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 $\sum _{n=0}^{\mathrm{\infty }}\frac{e^n}{\sqrt{n!}}$: 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: $L=\underset{n\to \mathrm{\infty }}{lim}\frac{e}{\sqrt{n+1}}=0$.
• 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.