/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Find the interval of convergence... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{x^{n}}{(n !)^{2}}$$

Short Answer

Expert verified
The interval of convergence is \(-\infty, \infty\).

Step by step solution

01

Identify the general form of the series

The given power series is \(\sum_{n=1}^{\infty} \frac{x^{n}}{(n !)^{2}}\).
02

Apply the Ratio Test

To determine the interval of convergence, use the Ratio Test. Compute the limit: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] where \(a_n = \frac{x^{n}}{(n !)^{2}}\).
03

Compute \(a_{n+1}\)

Substitute \(a_n = \frac{x^{n}}{(n !)^{2}}\) into the ratio test limit: \[ a_{n+1} = \frac{x^{n+1}}{((n+1)!)^{2}} \]
04

Form the ratio

Write the ratio of consecutive terms: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{x^{n+1}}{((n+1)!)^{2}}}{\frac{x^{n}}{(n !)^{2}}} \right| = \left| x \frac{(n!)^{2}}{((n+1)!)^{2}} \right| \].
05

Simplify the ratio

Simplify the expression: \[ \left| x \frac{(n!)^{2}}{((n+1)!)^{2}} \right| = \left| x \frac{(n!)^{2}}{(n+1)^{2}(n!)^{2}} \right| = \left| x \frac{1}{(n+1)^{2}} \right| \].
06

Evaluate the limit

Evaluate the limit as \(n \rightarrow \infty\): \[ L = \lim_{n \to \infty} \left| x \frac{1}{(n+1)^{2}} \right| = \left| x \right| \lim_{n \to \infty} \frac{1}{(n+1)^{2}} \]. Since \lim_{n \to \infty} \frac{1}{(n+1)^{2}} = 0, L = 0. The Ratio Test indicates that for any finite value of x, the series converges.
07

Determine the interval of convergence

Since the Ratio Test yields zero for all finite x, the series converges everywhere. Therefore, the interval of convergence is \(-\infty, \infty\).

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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 powerful tool used to determine the convergence of an infinite series. It's particularly useful for series where each term involves factorials, powers, or exponential functions.

The test works by examining the limit of the ratio of consecutive terms in a series. For a given series \(\sum a_n\), we define:
\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

In simple terms, if the limit L is less than 1, the series converges absolutely. If L is greater than 1, the series diverges. If L equals 1, the test is inconclusive, and we need to use other methods to check for convergence.

Let's apply this to our series \(\sum \frac{x^n}{(n!)^2}\). We'll compute the limit of the ratio of consecutive terms:
\[ a_{n+1} = \frac{x^{n+1}}{((n+1)!)^2} \]
\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| x \frac{(n!)^2}{((n+1)!)^2} \right| = \left| x \frac{(n!)^2}{(n+1)^2(n!)^2} \right| = \left| x \frac{1}{(n+1)^2} \right| \]

Evaluating the limit, we get:
\[ L = \lim_{n \to \infty} \left| x \frac{1}{(n+1)^2} \right| = |x| \lim_{n \to \infty} \frac{1}{(n+1)^2} = |x| \cdot 0 = 0 \]

Since the limit L is zero for all finite x, the Ratio Test confirms that our series converges for all x.
Power Series
A power series is a series of the form \(\sum a_n x^n \). It's essentially a polynomial with an infinite number of terms. These series are central to many areas in mathematics and are used to approximate functions.

For the power series \(\sum \frac{x^n}{(n!)^2}\), each term is determined by the formula \(a_n = \frac{1}{(n!)^2} \). The power series can converge or diverge, depending on the values of x.

Power series have an interval of convergence, which is the range of x values for which the series converges.

In this particular problem, applying the Ratio Test showed that our series converges for all x values. Thus, the interval of convergence is \(-\infty, \infty\).

It's important to remember the difference between absolute and conditional convergence. A series is absolutely convergent if the series of absolute values converges. If a power series converges absolutely within an interval, it usually converges conditionally at the endpoints of that interval.
Convergence in Mathematics
Convergence in mathematics refers to the idea that as we add more and more terms of a sequence or a series, the total sum approaches a specific value. There are several types of convergence, such as pointwise convergence, uniform convergence, and absolute and conditional convergence.

For infinite series, we primarily deal with absolute and conditional convergence:
  • *Absolute convergence:* A series \(\sum a_n\) converges absolutely if the series of absolute values \(\sum |a_n|\) converges.
  • *Conditional convergence:* A series \(\sum a_n\) converges conditionally if \(\sum a_n\) converges, but \(\sum |a_n|\) does not.
If a power series converges absolutely, it converges for all values within its radius of convergence.

In our problem, we used the Ratio Test to determine the limit of the ratio of consecutive terms, finding it to be zero for all finite x. This guarantees convergence.

It is also important to analyze the endpoints of the interval of convergence separately, since the behavior at the endpoints can differ from the rest of the interval. In this particular example, the interval of convergence is all real numbers, so we need not worry about endpoint behavior.

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Most popular questions from this chapter

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1-x^{2}}}$$

The series \(\sum_{n=1}^{\infty} 1 / n^{s}, s>1,\) is called the Riemann Zeta function, \(\zeta(s) .\) \((\text { a })\) you found \(\zeta(2)=\pi^{2} / 6 .\) When \(n\) is an even integer, these series can be summed exactly in terms of \(\pi .\) ) By computer or tables, find $$\text { (a) } \quad \zeta(4)=\sum_{n=1}^{\infty} \frac{1}{n^{4}}$$ $$\text { (b) } \quad \zeta(3)=\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$ $$\text { (c) } \quad \zeta\left(\frac{3}{2}\right)=\sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}}$$

(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.

Use series you know to show that: $$\ln 3+\frac{(\ln 3)^{2}}{2 !}+\frac{(\ln 3)^{3}}{3 !}+\cdots=2$$

Use the special comparison test to find whether the following series converge or diverge. $$\sum_{n=0}^{\infty} \frac{n(n+1)}{(n+2)^{2}(n+3)}$$

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