(In n)ª Problem 2. Use the Alternating Series Test to show that for any real number a, the series >(-1)"+! n n=1 converges. (Hint: You need to show that the sequence {(lnn)ª/n} converges to 0 for any a, and is eventually decreasing. Show that the function f(x) = (In x)ª/x is decreasing if æ is sufficiently large.)

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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(In n)ª
Problem 2. Use the Alternating Series Test to show that for any real number a, the series >(-–1)"+1
n
n=1
converges. (Hint: You need to show that the sequence {(ln n)ª/n} converges to 0 for any a, and is eventually
decreasing. Show that the function f(x) = (In x)ª /x is decreasing if x is sufficiently large.)
Transcribed Image Text:(In n)ª Problem 2. Use the Alternating Series Test to show that for any real number a, the series >(-–1)"+1 n n=1 converges. (Hint: You need to show that the sequence {(ln n)ª/n} converges to 0 for any a, and is eventually decreasing. Show that the function f(x) = (In x)ª /x is decreasing if x is sufficiently large.)
Expert Solution
Step 1

The given series  is :

n=1-1n+1ln nan 

Applying Alternative series test to the above series is given as:

The first condition is given as:

limnbn=0

Here, bn=ln nan    

limnbn=limnln nan =limnaln nn  =limnax1 =0      

This satisfies the first condition.

The second condition is given as :

bn must be a decreasing sequence.

bn=fnf'n=n×1n×a ln n-ln nan2=a ln n-ln nan2  

Thus, f'n0 

So, f is a decreasing sequence so fn+1<fn

Hence, bn+1<bn

This satisfies the second condition.

 

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