Use the preliminary test to decide whether the following series are divergent or require further testing. Careful: Do not say that a series is convergent; the preliminary test cannot decide this. $$\sum_{n=1}^{\infty} \frac{n !}{n !+1}$$

Short Answer

Expert verified
Since \( \lim_{{n \to \infty}} \frac{n!}{n!+1} = 1 \eq 0 \), the series is divergent.

Step by step solution

01

Understand the Preliminary Test

The preliminary test states that if \( \lim_{{n \to \infty}} a_n \eq 0 \) or does not exist, the series \( \sum a_n \) is divergent. This is important to remember as it will guide the process.
02

Define the General Term

Identify the general term \(a_n\) of the series. In this case, \(a_n = \frac{n!}{n!+1}\).
03

Compute the Limit

Evaluate the limit of the general term as \(n\) approaches infinity: \lim_{{n \to \infty}} \frac{n!}{n!+1}\.
04

Simplify the Limit Expression

Simplify the limit expression. Notice that \( \lim_{{n \to \infty}} \frac{n!}{n!+1} = \lim_{{n \to \infty}} \frac{1}{1 + \frac{1}{n!}} = 1 \) because \( n! \to \infty \) as \( n \to \infty \).
05

Apply the Preliminary Test

Using the preliminary test, since \( \lim_{{n \to \infty}} \frac{n!}{n!+1} = 1 \), and not 0, it implies the series \( \sum_{{n=1}}^{{\infty}} \frac{n!}{n!+1} \) is divergent.

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Key Concepts

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

Convergence Tests
To determine whether a series converges or diverges, mathematicians use various convergence tests. One of the simplest and often the first test we use is the Preliminary Test (or Divergence Test). This test states: If the limit of the nth term of a series does not approach zero as n approaches infinity, the series diverges.

In formula terms, if \(\text{lim}_{{n \rightarrow \infty}} a_n eq 0\) or the limit does not exist, then the series \(\text{ \sum a_n}\) diverges. This insight helps us quickly identify certain divergent series without using more complex tests.

Keep in mind there are more convergence tests like the Ratio Test, Root Test, and Integral Test, which help evaluate convergence in more challenging scenarios.
Limit of a Sequence
Understanding the limit of a sequence is crucial in determining series convergence. The limit is the value that the terms of a sequence approach as the index (usually n) goes to infinity. For instance, in our exercise, we needed to find the limit of the sequence given by the general term \(\frac{n!}{n!+1}\).

To compute the limit, observe that as n grows very large, the factorial \(n!\) also grows extremely large. This makes the term \( \frac{1}{n!} \) negligible in the denominator:

\[\lim_{{n \rightarrow \infty}} \frac{n!}{n! + 1} = \lim_{{n \rightarrow \infty}} \frac{1}{1 + \frac{1}{n!}} = 1 \]

This means the limit does not approach zero, which according to the preliminary test, means the series diverges.
Factorials in Series
Factorials are a common part of many series and sequences. The factorial of a number n, denoted as \(n!\), is the product of all positive integers up to n. Factorials grow extremely fast. For example, \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\).

In the provided exercise, we dealt with the general term \(\frac{n!}{n!+1}\). The rapid growth of factorials helps us simplify the limits involved.

Because \(n!\) grows so quickly as n increases, any additional constant in the denominator, like \(+1\), becomes insignificant, making the term simplify more easily when finding limits.

Understanding the properties of factorials will be very helpful in identifying how terms behave as n approaches infinity.
Evaluating Limits
Evaluating limits is a fundamental skill in calculus. It helps us understand the behavior of functions and sequences as the variable approaches a specific value, often infinity. When evaluating the limit of a fraction, as in our exercise, consider the behavior of the numerator and the denominator individually as n increases.

For \(\frac{n!}{n!+1}\), we simplify by finding the dominant terms in the numerator and the denominator. Here, as n approaches infinity, both the numerator \(n!\) and the denominator \(n! + 1\) grow, but since \(\frac{1}{n!}\) in \(n! + 1\) becomes very small, we focus on the term \( \frac{1}{1 + \frac{1}{n!}}\).

The limit evaluates to:

\[\lim_{{n \rightarrow \infty}} \frac{n!}{n! + 1} = 1 \]

Recognizing that the limit does not converge to zero means the series must diverge as per the preliminary test.

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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 the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16 ). $$\sum_{n=1}^{\infty} \frac{e^{n}}{e^{2 n}+9}$$

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 \(|x|<3\)

The energy of an electron at speed \(v\) in special relativity theory is \(m c^{2}\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) where \(m\) is the electron mass, and \(c\) is the speed of light. The factor \(m c^{2}\) is called the rest mass energy (energy when \(v=0\) ). Find two terms of the series expansion of \(\left(1-v^{2} / c^{2}\right)^{-1 / 2},\) and multiply by \(m c^{2}\) to get the energy at speed \(v\). What is the second term in the energy series? (If \(v / c\) is very small, the rest of the series can be neglected; this is true for everyday speeds.)

(a) Show that it is possible to stack a pile of identical books so that the top book is as far as you like to the right of the bottom book. Start at the top and each time place the pile already completed on top of another book so that the pile is just at the point of tipping. (In practice, of course, you can't let them overhang quite this much without having the stack topple. Try it with a deck of cards.) Find the distance from the right-hand end of each book to the right-hand end of the one beneath it. To find a general formula for this distance, consider the three forces acting on book \(n,\) and write the equation for the torque about its right-hand end. Show that the sum of these setbacks is a divergent series (proportional to the harmonic series). [See "Leaning Tower of The Physical Reviews," Am. J. Phys. 27, 121-122 (1959).] (b) By computer, find the sum of \(N\) terms of the harmonic series with \(N=25\) \(100,200,1000,10^{6}, 10^{100}\). (c) From the diagram in (a), you can see that with 5 books (count down from the top) the top book is completely to the right of the bottom book, that is, the overhang is slightly over one book. Use your series in (a) to verify this. Then using parts (a) and (b) and a computer as needed, find the number of books needed for an overhang of 2 books, 3 books, 10 books, 100 books.

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