/*! 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 9 Find the interval of convergence... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{x^{n}}{\ln (n+1)}$$

Short Answer

Expert verified
The interval of convergence is \([-1, 1)\).

Step by step solution

01

- Identify the General Term

The given series is \(\textstyle\sum_{n=1}^{\infty} \frac{x^{n}}{\ln (n+1)}\). Identify the general term, \(a_n\), which in this case is \(\frac{x^n}{\ln(n+1)}\).
02

- Apply the Ratio Test

The Ratio Test can help find the interval of convergence. Consider \(\textstyle\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\): \[\textstyle\lim_{n \to \infty} \left|\frac{\frac{x^{n+1}}{\ln(n+2)}}{\frac{x^n}{\ln(n+1)}}\right| = \lim_{n \to \infty} \left|x \cdot \frac{\ln(n+1)}{\ln(n+2)}\right| = |x| \lim_{n \to \infty} \[\frac{\ln(n+1)}{\ln(n+2)}\] = |x| \cdot 1 = |x|\].
03

- Determine Convergence from Ratio Test

By the Ratio Test, the series converges if \[|x| < 1\] and diverges if \[|x| > 1\]. Therefore, the open interval of convergence is \((-1, 1)\).
04

- Test Endpoints x = -1 and x = 1

Test the series at the endpoints.\At \[x = -1\]: The series becomes \(\textstyle\sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)}\), which is an alternating series. Since \(\textstyle\frac{1}{\ln(n+1)} \to 0\) as \(\textstyle n \to \infty\), the series converges by the Alternating Series Test.\At \[x = 1\]: The series becomes \(\textstyle\sum_{n=1}^{\infty} \frac{1}{\ln(n+1)}\), which diverges by the Integral Test as \(\textstyle\int_1^{\infty} \frac{1}{\ln(x+1)} \text{dx}\) diverges.
05

- Combine Results

Combine the results from the Ratio Test and the endpoint tests. The interval of convergence is \([-1, 1)\).

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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 common method for determining the convergence of a series. The Ratio Test involves taking the limit of the absolute value of the ratio of consecutive terms. For a series \(\textstyle\sum_{n=1}^{\infty} a_n\), the Ratio Test is used as follows: \[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L\]. If \[L < 1\], the series converges. If \[L > 1\], the series diverges. If \[L = 1\], the test is inconclusive.

In our problem, we applied the Ratio Test to the series \(\textstyle \sum_{n=1}^{\infty} \frac{x^n}{\ln(n+1)}\). By simplifying the limit, we found that \[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = |x|\]. This tells us that the series converges for \[|x| < 1\], giving an interval of \((-1, 1)\).
Alternating Series Test
The Alternating Series Test is useful for determining the convergence of series whose terms alternate in sign. Consider a series of the form \(\textstyle\sum(-1)^n b_n\), where \[b_n > 0\]. For the Alternating Series Test:
  • The series \[b_n\] is decreasing, and
  • \[b_n \to 0\] as \[n \to \infty\].
If both conditions are met, the series converges.

In the given exercise, testing the endpoints included checking for convergence of \(\textstyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)}\). Here, \(\textstyle \frac{1}{\ln(n+1)}\) decreases and goes to 0 as \(\textstyle n \to \infty\). Thus, by the Alternating Series Test, the series converges when \[x = -1\].
Integral Test
The Integral Test relates the convergence of a series to the convergence of an integral. For a series \(\textstyle\sum_{n=1}^{\infty} a_n\), if \[a_n\] is positive, continuous, and decreasing, we can use the Integral Test by comparing the series to an improper integral. The series converges if and only if \[\int_1^{\infty} f(x) \text{dx}\] converges, where \[f(n) = a_n\].

For \[x = 1\] in our problem, the series \(\textstyle \sum_{n=1}^{\infty} \frac{1}{\ln(n+1)}\) was evaluated using the Integral Test. We looked at \[\int_1^{\infty} \frac{1}{\ln(x+1)} \text{dx}\], which diverges. This shows that the series also diverges when \[x = 1\].
Analysis of Series
Understanding the convergence and divergence of series is crucial in calculus. Each test provides a different approach to analyzing series:
  • The Ratio Test is very versatile and often simplifies to a quick answer.
  • The Alternating Series Test relies on behavior of alternating terms and is simpler to check when applicable.
  • The Integral Test forms a bridge between understanding the sum of a series and integrals, often being useful for series with positive terms.


Combining these tests helps determine intervals of convergence. For \(\textstyle \sum_{n=1}^{\infty} \frac{x^n}{\ln(n+1)}\), the interval was found to be \([-1, 1)\) by using Ratio Test and end-point analysis with Alternating Series Test and Integral Test. This comprehensive understanding of series behaviors ensures a robust approach to many calculus problems.

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Most popular questions from this chapter

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$$

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n-1}}{2 n-1}$$

(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.

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.)

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

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