Determine If Integral Test Can Be Applied

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Bashyboy
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Homework Statement


I attached the infinite series the question provided as a file.


Homework Equations





The Attempt at a Solution


I deduced the general term to be ln(n)/n, so the infinite series would be written as [itex]\sum_{n=2}^{\infty} \frac{\ln(n)}{n}[/itex]

I took the derivative of the general term, and I found that the function is not decreasing on the entire interval I am summing on. The answer key, however, says the derivative is negative on the entire interval. Could someone please help me?
 
Attachments
  • Infinite series of fractions ln 2 over 2, plus ln 3 over 3, plus ln 4 over 4, plus ln 5 over 5, plus ln 6 over 6, plus continuing dots.
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Bashyboy said:

Homework Statement


I attached the infinite series the question provided as a file.

Homework Equations



The Attempt at a Solution


I deduced the general term to be ln(n)/n, so the infinite series would be written as [itex]\sum_{n=2}^{\infty} \frac{\ln(n)}{n}[/itex]

I took the derivative of the general term, and I found that the function is not decreasing on the entire interval I am summing on. The answer key, however, says the derivative is negative on the entire interval. Could someone please help me?

Well, what did you get for the derivative?
 
Bashyboy said:
(1-ln(x))/x^2
[itex]\displaystyle \frac{1-\ln(x)}{x^2}<0[/itex] for x > e .

So evaluate, [itex]\displaystyle \sum_{n=2}^{\infty} \frac{\ln(n)}{n}=\frac{\ln(2)}{2}+\sum_{n=3}^{ \infty} \frac{\ln(n)}{n}\ ,[/itex] since 3 > e.
 
Oh, so I can adjust the interval I am summing on?
 
I have another one I am working on. (The solution from the book is attached as a file). If the function that is comparable to the general term of this series is decreasing for values
x > e^1/2, why is one of the limits of integration x =1?
 
Attachments
  • Worked solution showing the series sum from n=1 to infinity of ln n over n squared; f(x) = ln x / x^2 with f'(x) = (1 - 2 ln x)/x^3, decreasing for x > e^(1/2) ≈ 1.6; the integral from 1 to infinity of ln x / x^2 dx equals 1, so the series converges by Theorem 9.10.
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Bashyboy said:
I have another one I am working on. (The solution from the book is attached as a file). If the function that is comparable to the general term of this series is decreasing for values
x > e^1/2, why is one of the limits of integration x =1?
So then, evaluate the integral from 2 to infinity.
 
So, is it technically improper to evaluate the integral from 1 to infinity?
 
Bashyboy said:
So, is it technically improper to evaluate the integral from 1 to infinity?
Quote the integral test, word for word.