Does SUM log(1+1/n)/n converge?

  • Thread starter Thread starter Treadstone 71
  • Start date Start date
  • Tags Tags
    Sum
Click For Summary

Homework Help Overview

The discussion revolves around the convergence of the series \(\sum \frac{1}{n}\log(1+\frac{1}{n})\), which falls under the subject area of series convergence tests in calculus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various convergence tests and comparisons, including the possibility of comparing the series to \(\frac{1}{n^2}\). There are discussions about the behavior of the logarithmic function and its bounds, as well as the application of Taylor series.

Discussion Status

The discussion includes attempts to apply comparison tests and Taylor series, with some participants questioning the bounds of the logarithmic function. While some guidance has been offered regarding comparisons, there is no explicit consensus on the approach or outcome yet.

Contextual Notes

Participants mention constraints such as the unbounded nature of the logarithmic function and the need for a more precise understanding of its behavior as \(n\) approaches infinity. There is also a reference to the Mercator series, which has not been covered in their studies.

Treadstone 71
Messages
275
Reaction score
0
Prove that [tex]\sum \frac{1}{n}\log(1+\frac{1}{n})[/tex] converges.

I tried a batch of tests, but none works. Last time Raabe's test was give to me upside down. I wonder if any other tests are give to me wrongly as well.
 
Physics news on Phys.org
Have you tried comparing it to 1/n^2?
 
I can't get it to work out
 
Post the details of your comparison...

What bounds do you know for log(1+x)?
 
It's unbounded.

I figured if [tex]\exp(\frac{1}{n}\log(1+\frac{1}{n}))=\exp(\frac{1}{n})+\exp(\frac{1}{n})/n \rightarrow 1[/tex] as [tex]n\rightarrow\infty[/tex], so at least the terms go to zero, although it's insufficient.

I could use the Mercator series, but we haven't seen it yet.
 
Last edited:
The first thing that was suggested was comparison- have you tried that?

The Taylor's series for ex is 1+ x + (1/2)x2+ ... so if x is positive, ex> 1+ x and x> log(1+ x) for all x. In particular, 1/n> log(1+ 1/n) for all n.
 
Treadstone 71 said:
It's unbounded.

It's bounded by a constant in the region you're concerned with. You are looking at log(1+x) where x=1/n, so 1<1+x<2.

A constant bound will not suffice here though, you need some indication of how fast this term is going to zero. Try the log(1+x)<x Halls mentioned, this was the one I was fishing for (one function can "bound" another).
 
I got it. Thanks for the help.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K