Test for convergence: $$\sum_{n=1}^{\infty} \frac{2^{n}}{n !}$$

Short Answer

Expert verified
The series \(\sum_{n=1}^{fty} \frac{2^{n}}{n !}\) converges by the Ratio Test.

Step by step solution

01

- Identify the Series

The given series is oindent\textbf{\(\sum_{n=1}^{\infty} \frac{2^{n}}{n !}\)}We need to test this series for convergence.
02

- Apply the Ratio Test

To determine if a series converges, apply the Ratio Test. The Ratio Test states that for a series \( \sum a_n \), consider the limit \[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If \(L < 1\), the series converges; if \(L > 1\), the series diverges; and if \(L = 1\), the test is inconclusive.
03

- Define Terms for the Ratio Test

In the given series \( a_n = \frac{2^n}{n!} \)We need to find: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} \right|\]
04

- Simplify the Ratio

Simplify the expression inside the limit: \[ \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2 \cdot 2^n}{(n+1) \cdot n!} \cdot \frac{n!}{2^n} = \frac{2 \cdot 2^n \cdot n!}{(n+1) \cdot n! \cdot 2^n} = \frac{2}{n+1} \]
05

- Evaluate the Limit

Evaluate the limit as \( n \) approaches infinity: \[ \lim_{n \to \infty} \frac{2}{n+1} = 0 \] Since 0 < 1, the ratio test confirms that the series converges.

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Key Concepts

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

Ratio Test
The Ratio Test is a handy tool in your mathematical toolkit when you want to determine if an infinite series converges or diverges. You start by examining the ratio of the terms in your series. For any series \( \sum a_n \, \), you need to find the limit:
\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]
Here are the rules you need to follow:
  • If \( L < 1 \), the series converges.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the Ratio Test doesn’t give a conclusion, and you’ll need another method.
For our series \( \sum_{{n=1}}^{{\forall}} \frac{2^n}{n!} \, \), applying the Ratio Test is straightforward because it simplifies easily, making calculations not only possible but manageable. Always remember, Ratio Test is a tool—it makes tough calculations easier!
Convergent Series
A Convergent Series is essentially a series whose terms approach a finite limit as you sum more and more of them. When we talk about an infinite series \( \sum a_n \, \) converging, we mean that the partial sums (i.e., the sum of the first \( n \) terms) get closer and closer to a specific value.
Here’s a quick example: if you keep adding smaller and smaller numbers that approach zero, the total sum will stabilize rather than explode to infinity.
This behavior is key in many areas of math and science because it helps to determine limits and behaviors of functions and formulas. So, recognizing whether a series converges is of utmost importance!
Limit Calculation
Evaluating limits is a critical part of calculus and analysis, particularly for determining series behavior. When applying the Ratio Test, calculating the limit accurately is paramount. This often involves simplifying complex expressions.
In our example series \( a_n = \frac{2^n}{n!} \,\), we found that:
\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \ = \ \lim_{{n \to \infty}} \frac{2}{n+1} = 0 \]
This calculation tells us that as \( n \) grows larger, the term \( \frac{2}{n+1} \) gets closer and closer to zero, which is less than one, confirming the series converges.
Performing these limit calculations involves some algebraic manipulation, and it's a skill worth practicing!
Factorial
The factorial, denoted \( n! \,\) is a mathematical operation where you multiply a positive integer \( n \) by all positive integers less than it. For example, \( 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \,\). It grows very quickly as \( n \) increases and is a crucial concept when working with series and permutations.
In our series \( \sum_{{n=1}}^{{\forall}} \frac{2^n}{n!} \, \), the factorial in the denominator plays a key role. It helps in decreasing the value of each term, contributing to the series’ convergence.
Factorials come up frequently in combinatorics, probability, and many areas of analysis, thus understanding them thoroughly is essential!

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Most popular questions from this chapter

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. \(\sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\) Method: Let \(y=x^{2}-1 .\) The power series \(\sum_{0}^{\infty} 8^{-n} y^{n}\) converges for \(|y|<8,\) so the original series converges for \(\left|x^{2}-1\right|<8,\) which means \(|x|<3\)

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