2
$\begingroup$

For a problem such as:

$ \sum_{n=1}^\infty \frac{lnn}{n^3}$

How do you tell if it is decreasing or increasing? I know that you can compute the derivative and test if it is less than or greater than zero, but the way my solutions manual does it, I'm puzzled by.

The derivative is:

$\frac{1-3lnx}{x^4} < 0$

Solving for x gives $x > e^\frac{1}{3}$.

I'm confused as to how that shows that the function is decreasing.

  • 2
    To nitpick, you are not really showing that the series is decreasing, but that the terms of the series are decreasing.2011-01-17

2 Answers 2

2

If a differentiable function has a negative derivative on an interval, then it is decreasing on that interval, a consequence of the mean value theorem. What you stated would therefore imply that the function $x\mapsto \frac{\ln x}{x^3}$ is decreasing on $(\sqrt[3]{e},\infty)$, so in particular it is decreasing on the integers greater than $1$.

0

Also, from the function you can work around the inequalities. You have

$\frac{1}{x^4}$ < $\frac{3lnx}{x^4}$

To prove this is true, since common denominator, 3lnx has to be greater than 1 for some interval. Which is true by the graph of 3lnx that increases above 1