   Chapter 11.3, Problem 32E

Chapter
Section
Textbook Problem

# Find the values of p for which the series is convergent.32. ∑ n = 1 ∞ ln n n p

To determine

To find: The value of p if the series is convergent.

Explanation

Given:

The series is n=1an=n=1lnnnp.

Definition used:

The improper integral abf(x)dx is convergent if the limit exists, otherwise it is divergent.

Result used:

(1) If the function f(x) is continuous, positive and decreasing on [1,) and let an=f(n), then the series n=1an is convergent if and only if the improper integral 1f(x)dx is convergent; otherwise, it is divergent.

(2) The function f(x) is decreasing function if f(x)<0.

Chain rule: d[f(x)]ndx=n[f(x)]n1f(x)

Product rule: ddx[f(x)g(x)]=dfdxg(x)+f(x)dgdx

Integration by parts: If g(x) and h(x) are differentiable, then g(x)h(x)dx=g(x)h(x)h(x)g(x)dx.

Calculation:

The given function lnxxp is not decreasing if p0.

Consider the function lnxxp when p>0.

The derivative of the function is obtained as follows:

f(x)=xpddx(lnx)+(lnx)ddx(xp)    [ddx[f(x)g(x)]=dfdxg(x)+f(x)dgdx]=xp(1x)+(lnx)(pxp1)         [d[f(x)]ndx=n[f(x)]n1f(x)]=xp1(plnx1)

Since f(x)<0 for x>e1p, the given function is decreasing by using the Result (2).

Clearly, the function f(x) is continuous, positive and decreasing on [1,).

Use the Result (1), the series is convergent if the improper integral lnxxp is convergent.

By the definition, the improper integral is convergent if the limit exists.

Compute 1lnxxpdx as shown below.

1lnxxpdx=limt[1tlnxxpdx] (1)

Obtain the integral 1tlnxxpdx.

Consider the indefinite integral lnxxpdx by using Integration by parts Rule,

Substitute g(x)=lnx and h(x)=1xp,

lnxxpdx=lnx(xp+1p+1)(xp+1p+1)1xdx=lnx(xp+1p+1)xpp+1dx=lnx(xp+1p+1)1p+1(xp+1p+1)=lnx(x1p)1p

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