# Homework Help: Convergence or divergence of log series

1. Jul 11, 2007

### Rosey24

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

I'm supposed to evaluate the following series or show if it diverges:

SUM (sigma) log [(x+1)/x]

2. Relevant equations

Drawing a blank...

3. The attempt at a solution

I'm unsure how to start this. We've gone over all sorts of tests for convergence (ratio, comparison, p-series, etc), but I'm not sure what to do with the log function. A push in the right direction would be appreciated!

Thanks.

2. Jul 11, 2007

### CompuChip

What are you summing over (I assume x, though x usually denotes a continuous variables, and indices like k or n are commonly used for discrete summation) and what are your bounds (if x = 0 is included in the summation you immediately see that it diverges, but if $x \ge 1$ this is not immediately obvious).
In other words, please be a little more specific.

3. Jul 11, 2007

### Rosey24

I recalled this incorrectly, it's log[(n+1)/n] and no bounds were given, though I would assume it's 1 to infinity as zero would diverge.

4. Jul 11, 2007

### bob1182006

ok n=1, n->infinity, before you think about the log, what happens to the (n+1)/n ? figure that out first then you can figure out if the log of that converges/diferges

5. Jul 11, 2007

### Rosey24

as n goes to infinity, (n+1)n will converge to 1, right? and log (1) is zero.

6. Jul 11, 2007

### bob1182006

Right, but that just tells you the series CAN converge, for this problem I'm thinking L'Hospital's rule due to the log.
A nice series to compare with is the Harmonic Series, 1/x, since (n+1)/n = 1 +1/n.

Edit: sry comparing with 1/x doesn't work >< Just woke up and automatically did L'Hospital's rule without checking if the comparison gave an indeterminate form.

EditEdit: ok it does work >.> it gives you 0/0 lol

Last edited: Jul 11, 2007
7. Jul 11, 2007

### quasar987

I suggest obtaining a lower bound on log [(x+1)/x] = log [1+1/x] by use of a Taylor expansion, and then comparing the sum of logs with the sum of that lower bound.

8. Jul 11, 2007

### NateTG

It's pretty easy to find a closed form for the partial sum.

Big stinking giveaway hint:
$$\log \frac{a}{b}=\log{a}-\log{b}$$