Convergence and Divergence of a series

[mohabitar]

The series from n=1 to infinity log(n/(n+1)). This was on my quiz, which I got wrong. Here's what I did:
lim n-->infinity of log(n/(n+1))
so then that becomes: log(lim n-->infinity n/(n+1))
which becomes the log1, which is 0, so it converges.

Whats wrong with my steps?

 

[rakalakalili]

What you showed was that the sequence converges to 0, not the series. The sequence has to converge to 0 for the series to converge, but it does not guarantee that it does. A good example of this is the harmonic series
$$\sum_{k=1}^{\infty} \frac{1){k}$$
The limit as k goes to infinity of 1/k is zero, but the series still diverges.

 

[Mark44]

If you have a series $\sum a_n$, and lim a_n is not 0, or the limit doesn't exist, then you know that your series diverges.

If lim a_n = 0, then you really can't say much at all about your series. That's what all the tests (comparison, ratio, root, integral, limit comparison, etc.) are about.

 

[mohabitar]

I thought what I did was a test?? Ahhh this stuff is so confusing! So many times in my book it said if the lim of an=0, then it converges. I don't get what's going on.

 

[Mark44]

I thought what I did was a test?? Ahhh this stuff is so confusing! So many times in my book it said if the lim of an=0, then it converges. I don't get what's going on.

I suspect that you are confusing a sequence, {a_n}, with a series, $\sum a_n$. If lim a_n = 0 (or any specific number), the sequence converges, but nothing can be said about the series $\sum a_n$.

The harmonic series that rakalakalili gave and this series $$\sum_{k=1}^{\infty} \frac{1}{k^2}$$
are such that lim a_n = 0, but the harmonic series diverges and the other series converges.