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Chapter 1: Problem 4

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

Short Answer

Expert verified
The series converges by the Ratio Test.

Step by step solution

01

Identify the Series

The series given is \(\sum_{n=1}^{\infty} \frac{2^{n}}{n !}\). Recognize that each term of the series is in the form of \( \frac{a^n}{n!} \) with \( a = 2 \).
02

Apply the Ratio Test

Use the Ratio Test to determine the convergence of the series. The Ratio Test states that for a series \( \sum_{n=1}^{\infty} a_n \), if \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] \( L < 1 \) then the series converges. Here, \( a_n = \frac{2^n}{n!} \).
03

Compute the Ratio

Calculate \(\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} \right| = \left| \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} \right| = \left| \frac{2 \cdot 2^n \cdot n!}{2^n \cdot (n+1)n!} \right| = \left| \frac{2}{n+1} \right|\).
04

Evaluate the Limit

Find the limit \(L = \lim_{n \to \infty} \left| \frac{2}{n+1} \right| = \lim_{n \to \infty} \frac{2}{n+1} = 0\). Since \(0 < 1\), the series converges.

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Key Concepts

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

Mathematical Series
In mathematics, a series is the sum of the terms of a sequence. For instance, the series \(\sum_{n=1}^{\infty} a_n\)\ represents the infinite sum of the sequence \{a_n\}. A series can be finite or infinite and understanding their behavior is key. Infinite series, like the given problem \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\),\ involve an infinite number of terms added together.
Ratio Test
The Ratio Test is a common method to determine whether an infinite series converges or diverges. It's especially useful for series where terms involve factorials or exponentials. To apply the Ratio Test, we calculate the limit: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \.\] If \L \lt 1\, the series converges; if \L \gt 1\, it diverges. If \L = 1\, the test is inconclusive.
Infinite Series Convergence
Determining the convergence of an infinite series helps us understand its sum approaches a finite value. For instance, the series \(\sum_{n=1}^{\infty} \frac{2^{n}}{n!}\)\ converges because the Ratio Test limit \(\lim_{n \to \infty} \frac{2}{n+1} = 0 \lt 1\).\
In these cases, as \ becomes very large, the ratio between consecutive terms approaches zero, indicating that the series sum reaches a limit.

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Most popular questions from this chapter

Write the following series in the abbreviated \(\sum\) form. $$ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots $$

Use the methods of this section to find the first few terms of the Maclaurin series for each of the following functions. $$ \ln \left(1+x e^{x}\right) $$

Use power series to evaluate the function at the given point. Observe in each case that, even with a hand calculator available, it is easier or necessary to use series. \(e^{-x^{2} / 2}-1-\ln \cos x\) at \(x=0.0024\)

Find the sum of the series $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots $$ Hint : \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\) Show that the remainder after \(n\) terms is \(1 /(n+1)\). Hence show that about 200 terms are needed for two decimal place accuracy. Compare the remainder with the value of the 200th term and so show that in computation using series of positive terms the value of the first omitted term may be a completely unreliable estimate of the error.

\(\sum_{n=1}^{\infty} \frac{n^{5}}{5^{n}}\)
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