# why sum 1/x is diverge

• Mar 31st 2009, 08:31 PM
joseph0177
why sum 1/x is diverge
I know by intergal test

1/x => ln x which is diverge

but lim 1/x = 0 as x -> infinity
which satisify the theorem in my test , the kth term of a convergent series tends to 0.

if sum ak for k from 0 to infinity converges, the ak -> 0 as k -> infinity
• Mar 31st 2009, 09:08 PM
o_O
The theorem says: If $\displaystyle \sum_{n=0}^{\infty} a_n$ converges, then $\displaystyle \lim_{n \to \infty} a_n = 0$.

This does NOT mean that it is true the other way around!

For example, if I told you that $\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}$ converges, then according to the above, without a doubt you know that $\displaystyle \lim_{n \to \infty} \frac{n!}{n^n} = 0$.

However, if all I told you was $\displaystyle \lim_{n \to \infty} \frac{n!}{n^n} = 0$, the theorem doesn't say anything about the statement backwards so we don't know if it converges or not and you would have to use some test to prove that it does.

Do you get it? Essentially, the theorem is the statement in the form: If $\displaystyle A$, then $\displaystyle B$.

Just because $\displaystyle B$ is occurred, doesn't mean $\displaystyle A$ did.