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

  • #1
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.
 

Answers and Replies

  • #2
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Have you tried comparing it to 1/n^2?
 
  • #3
I can't get it to work out
 
  • #4
shmoe
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Post the details of your comparison...

What bounds do you know for log(1+x)?
 
  • #5
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:
  • #6
HallsofIvy
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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.
 
  • #7
shmoe
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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).
 
  • #8
I got it. Thanks for the help.
 

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