Convergent Series: ln(n/(n+1))

[mickellowery]

Homework Statement

∑ ln((n)/(n+1)) I was assuming this would be $$\infty$$/$$\infty$$
and if I divide through by n it gives me 1/1 or 1 so would this just be divergent?

Homework Equations

The Attempt at a Solution

 

[physicsman2]

So is this a specific test you're trying to use?

The limit goes to 0. Factor out an n from n/(n+1) to get [n(1)]/[n(1 + 1/n)].

Cancel the n to get 1/[1+ 1/n]. When you apply the limit, you should get ln(1/1) which just equals 0.

The only problem with this is I'm not sure if you have to change the n's to x's to make it a continuous function in order to cancel the n's, but either way that's how you get the answer.

 

[Mark44]

Homework Statement

∑ ln((n)/(n+1)) I was assuming this would be $$\infty$$/$$\infty$$
and if I divide through by n it gives me 1/1 or 1 so would this just be divergent?

Homework Equations

The Attempt at a Solution

$\infty/\infty$ is meaningless as a final answer to anything (it is called indeterminate), and you're ignoring the fact that the general term in your series is ln(n/(n + 1)). As n gets large, n/(n + 1) --> 1, so your general term --> 0. This tells you precisely nothing about your series, so you're going to need to do something else. What other tests do you know?

 

[physicsman2]

$\infty/\infty$ is meaningless as a final answer to anything (it is called indeterminate), and you're ignoring the fact that the general term in your series is ln(n/(n + 1)). As n gets large, n/(n + 1) --> 1, so your general term --> 0. This tells you precisely nothing about your series, so you're going to need to do something else. What other tests do you know?

Hey Mark, do you have to make it a continuous function to cancel the n's after factoring them out, the way I did it?

 

[mickellowery]

OK so I'm trying to use the Integral test and I've gotten to lim$$\int$$ln(x/x+1)dx but now I'm not sure how to proceed. I was wondering if I could divide this into two separate integrals. I'm a little confused.

 

[Mark44]

Hey Mark, do you have to make it a continuous function to cancel the n's after factoring them out, the way I did it?

I'm not sure I understand your question, but I'll take a stab at it. lim [ln(f(x))] = ln[lim (f(x))] as long as f is continuous. In this case f(x) = x/(x + 1), which is continuous for x > - 1. For the series in this problem, it's not shown, but I suspect that n ranges from 1 to infinity.

 

[System]

Apply one of the logarithmic laws, and you will face a telescoping series