Converge or diverge?

[KataKoniK]

I am a bit confused here of whether the following functions diverge or converge.

\sum \ln x \backslash x

I used teh Integral test and got an answer showing the series is divergent where

lim ((ln b)^2) / 2 = infinity
b-> infinity

Is this correct? I graphed the equation and saw an answer such that the equation converged to 0.

\sum 2/(x (lnx)^2)

I don't think I did this right either. Using the integral test I got

lim -2/(ln b) + 2 / ln(2) = 2 / ln(2)
b-> infinity
So it converges to 2 / ln(2)

Am I allowed to assume that -2/(ln b) -> 0 as b-> infinity?

[matt grime]

presumably you're summing over x in N.

The first diverges by the comparson test:

log(n)/n >1/n

and sum 1/n diverges.

The second I think does converge by the integral test, but the limits of the sum and the integral are NOT the same, it is merely an existence test, it does not tell you what it is.

incidentally, what equations are you referring to when you talk about graphing them?

[KataKoniK]

I might be wrong here, but for the first equation (ln x / x) I graphed it into the calculator and saw the graph getting closer and closer to 0 as x -> infinity.

Thanks for your help.

[shmoe]

I might be wrong here, but for the first equation (ln x / x) I graphed it into the calculator and saw the graph getting closer and closer to 0 as x -> infinity.

log(x)/x does go to zero as x goes to infinity, but the series \sum(\log n)/n still diverges. Don't mix up the limit of the terms with the limit of their partial sums.

Or did you mean you graphed \log(x)^2/2 and found it was going to zero as x whent to infinity? If so, check again. What is log(x) doing as x grows?