/*! 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 42 Use any method to determine whet... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use any method to determine whether the series converges. $$\sum_{k=1}^{\infty}\left(\frac{k}{k+1}\right)^{k^{2}}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Analyze the General Term

The general term of the series is given by \( a_k = \left(\frac{k}{k+1}\right)^{k^2} \). To determine convergence, we need to analyze the behavior of these terms as \( k \to \infty \). The expression can be rewritten with an exponential function: \[a_k = e^{k^2 \ln\left(\frac{k}{k+1}\right)}.\]
02

Simplify the Logarithmic Expression

We approximate \( \ln\left(\frac{k}{k+1}\right) \). Using the property \( \ln\left(1 - x \right) \approx -x \) for small \( x \), where \( x = \frac{1}{k+1} \), we find:\[\ln\left(\frac{k}{k+1}\right) = \ln\left(1 - \frac{1}{k+1}\right) \approx -\frac{1}{k+1}.\]
03

Apply Simplification to Exponent

Substitute the approximation into the exponent of \( a_k \):\[a_k = e^{k^2 (-\frac{1}{k+1})} = e^{-\frac{k^2}{k+1}}.\]As \( k \to \infty \), the term \(-\frac{k^2}{k+1}\) simplifies to \(-k\), since \( \frac{k^2}{k+1} \approx k \) for large \( k \).
04

Evaluate the Limit of the General Term

Now evaluate the limit:\[a_k = e^{-k} \to 0 \text{ as } k \to \infty.\]Although \( a_k \to 0 \), we need to assess it to determine if it's a valid series for convergence tests.
05

Use the Ratio Test

To decide on absolute convergence, apply the ratio test:\[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|,\]where:\[a_{k+1} = \left(\frac{k+1}{k+2}\right)^{(k+1)^2}.\]After simplification and substitution, we find:\[L = e^1 > 1.\]
06

Conclusion on Convergence

Since \( L = e > 1 \), the series \( \sum_{k=1}^{\infty}\left(\frac{k}{k+1}\right)^{k^{2}} \) diverges by the ratio test. The individual terms decreasing to zero is necessary, but not sufficient for convergence.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Ratio Test
The ratio test is a useful technique for analyzing the convergence of sequences and series. It simplifies the process of determining whether a series converges absolutely. To apply the ratio test, start by considering the ratio of consecutive terms in the series. For a series \( \sum a_k \), calculate the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). The outcome of the ratio test depends on the value of this limit:
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In our exercise, the term \( a_k = \left(\frac{k}{k+1}\right)^{k^2} \) was analyzed. By using the ratio test, the limit \( L \) was calculated as \( e > 1 \). This clearly indicates divergence. The ratio test provides a straightforward approach especially when the series terms involve factorials or exponentials that lend themselves well to simplification.
Divergent Series
A divergent series is one in which the partial sums do not settle at a finite number. Essentially, as you keep adding more terms, the sum grows indefinitely rather than stabilizing. When analyzing series, recognizing divergence is crucial since it implies that the series does not have a meaningful sum in the traditional sense.In our original exercise, we derived the expression \( a_k = \left(\frac{k}{k+1}\right)^{k^2} \), and despite the terms themselves becoming very small, they do not ensure convergence. The series ultimately diverges because, based on the ratio test, the calculated limit \( L = e > 1 \) indicated that the series expansion grows larger without bound. While terms approaching zero might hint at convergence, it's not conclusive, and thus tests like the ratio test are applied for certainty.
Exponential Functions
Exponential functions are fundamental in various branches of mathematics and science. They describe processes that change at rates proportional to their current value, yielding forms such as \( e^x \). When dealing with series, exponentials often come up, especially when expressing terms that involve growth or decay, like in our exercise: \( a_k = e^{-\frac{k^2}{k+1}} \).Exponential functions are characterized by their continuous and smooth curves and have interesting properties:
  • The base \( e \) is approximately 2.718, known as Euler's number.
  • Derivatives of \( e^x \) remain \( e^x \), displaying unique self-similarity.
  • Exponentials can approximate large growths and declines, making them invaluable in both mathematical theory and practical applications.
In our specific case, we saw how exponential notation facilitated the analysis of the term \( \left(\frac{k}{k+1}\right)^{k^2} \) by expressing it as \( e^{k^2 \ln\left(\frac{k}{k+1}\right)} \). This transformation enabled the application of approximation techniques and convergence tests.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the Taylor polynomials of orders \(n=0,1,2,3\) and 4 about \(x=x_{0},\) and then find the \(n\) th Taylor polynomial for the function in sigma notation. $$\frac{1}{x} ; x_{0}=-1$$

Find the first four distinct Taylor polynomials about \(x=x_{0},\) and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. $$f(x)=\sin x ; x_{0}=\pi / 2$$

Exercise will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises. It was stated in Exercise 35 that $$\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}$$ (a) Let \(s_{n}\) be the \(n\) th partial sum of the series above. Show that $$ s_{n}+\frac{1}{3(n+1)^{3}} < \frac{\pi^{4}}{90} < s_{n}+\frac{1}{3 n^{3}} $$ (b) We can use a partial sum of the series to approximate \(\pi^{4} / 90\) to three decimal-place accuracy by capturing the sum of the series in an interval of length 0.001 (or less). Find the smallest value of \(n\) such that the interval containing \(\pi^{4} / 90\) in part (a) has a length of 0.001 or less. (c) Approximate \(\pi^{4} / 90\) to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of \(\pi^{4} / 90\).

Use a CAS to confirm that $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}$$ and then use these results in each part to find the sum of the series. $${(a)}\sum_{k=1}^{\infty} \frac{3 k^{2}-1}{k^{4}}$$ $${(b)}\sum_{k=3}^{\infty} \frac{1}{k^{2}}$$ $${(c)}\sum_{k=2}^{\infty} \frac{1}{(k-1)^{4}}$$

Find the Taylor polynomials of orders \(n=0,1,2,3\) and 4 about \(x=x_{0},\) and then find the \(n\) th Taylor polynomial for the function in sigma notation. $$e^{-x} ; x_{0}=\ln 2$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.