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Q32E

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Found in: Page 453

### Essential Calculus: Early Transcendentals

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

# 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 $$p \le 1$$.

See the step by step solution

### Step by Step Solution

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 $$p \le 1$$.