# Convergent Series

1. May 4, 2010

### mickellowery

1. The problem statement, all variables and given/known data
∑ 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?

2. Relevant equations

3. The attempt at a solution

2. May 4, 2010

### 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.

3. May 4, 2010

### Staff: Mentor

$\infty/\infty$ 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?

4. May 4, 2010

### physicsman2

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?

5. May 4, 2010

### 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.

6. May 4, 2010

### Staff: Mentor

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.

7. May 4, 2010

### System

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

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