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Q32E

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Essential Calculus: Early Transcendentals
Found in: Page 453
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find the values of p for which the series is \(\sum\limits_{n = 1}^\infty {\frac{{lnn}}{{{n^p}}}} \)convergent.

The series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) is convergent for \(p > 1\) and divergent for \(p \le 1\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

About Comparison test:-

Consider the series, \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \).

According to the comparison test, suppose that \(\sum {{a_n}} \)and \(\sum {{b_n}} \) are series with positive terms.

(i) If \(\sum {{b_n}} \)is convergent and \({a_n} \le {b_n}\forall n\)then \(\sum {{a_n}} \)is also convergent.

(ii) If \(\sum {{b_n}} \)I is divergent and \({a_n} \ge {b_n}\forall n\) then \(\sum {{a_n}} \)is also divergent.

(iii) For \(n \ge 3,\ln n > 1\) so for \(n \ge 3,\frac{{\ln n}}{{{n^p}}} > \frac{1}{{{n^p}}}\).

(iv) Apply comparison test, since only the eventual behavior of series determines convergence rather than the first several terms only.

The p-series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) is convergent if \(p > 1\) and divergent if\(p \le 1\).

Consider the case where \(p \le 1\):-

The series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) and \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) are series of positive terms. As the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) is a p-series with \(p \le 1\). Therefore, the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) is convergent.

The terms of \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) are bigger than the terms of \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \), \(\frac{{\ln n}}{{{n^p}}} > \frac{1}{{{n^p}}}\) for \(n \ge 3\).

\(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \)must also diverge. So, the series diverges for \(p \le 1\).

Consider the case where \(p > 1\):-

If \(f(x) = \frac{{\ln x}}{{{x^p}}},\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} = f(n)\).also, f is eventually decreasing. So, use the integral test. We demonstrate that f is eventually decreasing by taking its derivative with quotient rule:

\(f'(x) = \frac{{{x^p}\frac{1}{x} - \ln x.p{x^{p - 1}}}}{{{x^{2p}}}} = \frac{{{x^{p - 1}} - \ln xp{x^{p - 1}}}}{{{x^{2p}}}}\)

\( = \frac{{{x^{p - 1}}(1 - p\ln x)}}{{{x^{2p}}}}\)

The factors \({x^{p - 1}}\) and \({x^{2p}}\)are positive for positive values of x. Also \(p > 1\), for \(x \ge 3\). So, \(x \ge 3\), \(p\ln x > 1\), giving that \(0 < 1 - p\ln x\). Then for \(x \ge 3\), \(\frac{{{x^{p - 1}}(1 - p\ln x)}}{{{x^{2p}}}}\) is the product of one negative and two positive factors, therefore, negative. The integral test still applies because it is eventual behaviour of integral & series that matters for convergence, not its value from \(1 \le x \le 3\). So, f has a decreasing property that allow us to use the integral test.\(\)

Evaluation of integral:-

For evaluation of \(\int\limits_1^\infty {f(x)dx} \), first we compute the indefinite integral: \(\int {f(x)dx} = \int {\frac{{\ln x}}{{{x^p}}}dx} \).

Use integration by parts:-

Let \(u = \ln x,{\rm{ }}du = {x^{ - p}}dx\).

Then, \(du = \frac{1}{x}dx{\rm{, }}u = \frac{1}{{ - p + 1}}{x^{ - p + 1}}\).

Note that it’s important that \(p > 1\)so \( - p + 1 \ne 0\).

By the integration by parts, we have

\(\int {\frac{{\ln x}}{{{x^p}}}dx = \frac{{\ln x}}{{{x^p}}}{x^{ - p + 1}}} - \int {\frac{1}{{ - p + 1}}{x^{ - p + 1}}\frac{1}{x}dx} \)

\( = \frac{{\ln x}}{{ - p + 1}}.{x^{p + 1}} - \frac{1}{{{{( - p + 1)}^2}}}{x^{ - p + 1}}\)

\( = \frac{{\ln x - \frac{1}{{ - p + 1}}}}{{( - p + 1){x^{p - 1}}}}\)

Take the improper integral

\(\int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} = \mathop {\lim }\limits_{t \to \infty } \int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} \)

\( = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{\ln x - \frac{1}{{ - p + 1}}}}{{( - p + 1){x^{p - 1}}}}} \right]_1^t\)

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln t - \frac{1}{{ - p + 1}}}}{{( - p + 1){x^{p - 1}}}} + \frac{1}{{{{( - p + 1)}^2}}}\)

The term \(\ln t \to \infty \) and since \(p > 1\), also. So, the \(\frac{1}{{ - p + 1}}\) has the indefinite form\(\frac{\infty }{{ - \infty }}\) because \(\frac{1}{{ - p + 1}}\)in the numerator and \( - p + 1\) are constants.

using L’Hospital’s rule:-

Apply L’Hospital’s rule, to calculate:

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln t - \frac{1}{{ - p + 1}}}}{{( - p + 1){t^{p - 1}}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{(1/t)}}{{\frac{{ - p + 1}}{p}{t^p}}}\)

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{(1/t)}}{{\frac{{ - p + 1}}{p}{t^{p + 1}}}} = 0\) [∵ p + 1 > 2]

So, above the limit converges to zero, and thus the main limit convergent.

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln t - \frac{1}{{ - p + 1}}}}{{( - p + 1){t^{p - 1}}}} + \frac{1}{{{{( - p + 1)}^2}}} = \frac{1}{{{{( - p + 1)}^2}}}\)

The integral \(\int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} \) converges.

By integral test, \(\int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} \) converges if & only if the series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \)converges. Thus, the series is convergent.

Therefore, \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) is convergent for \(p > 1\) and divergent for \(p \le 1\).

Most popular questions for Math Textbooks

\({\bf{37 - 40}}\) Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?

\({{\bf{a}}_{\bf{n}}}{\bf{ = n( - 1}}{{\bf{)}}^{\bf{n}}}\)

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({\bf{\{ 0,1,0,0,1,0,0,0,1, \ldots \ldots \ldots \ldots \ldots \} }}\)

(a) Find the partial sum S10of the series\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} \) . Use Exercise 33(a) to estimate the error in using S10 as an approximation to the sum of series.

(b) Use exercise 33(b) with n=10to give an improved estimate of the sum.

(c) Find a value of n so that \({{\bf{S}}_{\bf{n}}}\)is within 0.00001 of the sum.

Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent?

\(\sum\limits_{n = 1}^\infty {\frac{1}{{\ln (n + 1)}}} \)

(a) Let, \({a_1} = a\), \({a_2} = f\left( a \right)\), \({a_3} = f\left( {{a_2}} \right) = f\left( {f\left( a \right)} \right)\), . . . ,\({a_{n + 1}} = f\left( {{a_n}} \right)\), where \(f\)is a continuous function. If\(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\), show that \(f\left( L \right) = L\).

(b) Illustrate part (a) by taking \(f\left( x \right) = \cos x\), \(a = 1\), andestimating the value of \(L\)to five decimal places.
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