Determine If Integral Test Can Be Applied

1. Oct 6, 2012

Bashyboy

1. The problem statement, all variables and given/known data
I attached the infinite series the question provided as a file.

2. Relevant equations

3. The attempt at a solution
I deduced the general term to be ln(n)/n, so the infinite series would be written as $\sum_{n=2}^{\infty} \frac{\ln(n)}{n}$

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?

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2. Oct 6, 2012

SammyS

Staff Emeritus
Well, what did you get for the derivative?

3. Oct 6, 2012

Bashyboy

(1-ln(x))/x^2

4. Oct 6, 2012

SammyS

Staff Emeritus
$\displaystyle \frac{1-\ln(x)}{x^2}<0$ for x > e .

So evaluate, $\displaystyle \sum_{n=2}^{\infty} \frac{\ln(n)}{n}=\frac{\ln(2)}{2}+\sum_{n=3}^{ \infty} \frac{\ln(n)}{n}\ ,$ since 3 > e.

5. Oct 6, 2012

Bashyboy

Oh, so I can adjust the interval I am summing on?

6. Oct 6, 2012

Bashyboy

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?

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7. Oct 6, 2012

SammyS

Staff Emeritus
So then, evaluate the integral from 2 to infinity.

8. Oct 7, 2012

Bashyboy

So, is it technically improper to evaluate the integral from 1 to infinity?

9. Oct 7, 2012

SammyS

Staff Emeritus
Quote the integral test, word for word.