Does the series SUM log(1+1/n) converge or diverge?

[happyg1]

Hello,
Here's the question:
Does the series SUM log(1+1/n) converge or diverge?

I wrote out the nth partial sums like this:

log(1+1) + log(1+1/2) + log(1+1/3)+...+log(1+1/n)

It looks to me like the limit of the thing inside the parentheses goes to 1 as n goes to infinity, making the limit of the entire thing 0. So I say it converges.

One of my classmates says that SUM 1/n diverges, so this one does too. I can't disagree with him, but I fail to see the relevance.

I'm confused. Any clarification will be appreciated.
CC

 

[shmoe]

It looks to me like the limit of the thing inside the parentheses goes to 1 as n goes to infinity, making the limit of the entire thing 0. So I say it converges.

The terms going to zero is a necessary but not sufficient condition for convergence of the series (1/n is your basic counterexample).

One of my classmates says that SUM 1/n diverges, so this one does too. I can't disagree with him, but I fail to see the relevance.

They probably have the limit comparison test in mind. You can also look at the integral comparison test, you can find an antiderivative of log(1+1/n) easily enough.

 

[Ali 2]

The series diverges ..
Take the sequance of partial sums ..
S_n = \sum_{k=1} ^n\log \left( 1 + \frac 1k \right) = \sum_{k=1} ^n \log \left ( \frac { k+1} { k} \right ) <br /> = \sum_{k=1} ^n \log (k+1) - \log (k) \mbox{ ( Telescoping series }

= (\log 2 - \log 1) + (\log 3 - \log 2) +... + (\log n - \log (n-1) ) + (\log (n+1) - \log n)

= - \log 1 + \log ( n+1 ) = \log (n+1)
\lim _ { n \rightarrow \infty } S_n = \lim _ { n \rightarrow \infty } \log ( n+1) = \infty <br /> \Longrightarrow \sum_{n=1} ^ {\infty}\log \left( 1 + \frac 1n \right) \mbox{ diverges }