# Comparision Test.

1. Oct 31, 2007

### azatkgz

1. The problem statement, all variables and given/known data
Determine whether the series $$\sum_{n=2}^{\infty}a_n$$ absolutely convergent,conditionally convergent or divergent if

$$a_n=\frac{1}{ln(n!)}$$

3. The attempt at a solution

To solve this problem I used Comparision Test,so

$$\frac{1}{ln(n!)}<\frac{1}{n}$$ for all n>6

So I concluded that this series has finite limit

Last edited: Oct 31, 2007
2. Oct 31, 2007

### sutupidmath

from what u have done, you defenitely cannot conclude whether this series converges or diverges, since the series of 1/n diverges..

Here is how i think u should go about solving this.

since n!<n^n, and using the fact that ln(n) is monotonically increasing we have that

ln(n!)<ln (n^n)=n*ln(n) , or

1/n*ln(n)<1/ln(n!) and now since the series

$$\sum_{n=2}^{\infty} 1/(n ln(n))$$ diverges we can conclude that also the series 1/ln(n!) when n goes from 2 to infinity diverges.

3. Oct 31, 2007

### sutupidmath

Do u know how to show now that the latter series that i wrote

$$\sum_{n=2}^{\infty} \frac {1}{n ln(n)}$$

diverges???

4. Oct 31, 2007

### azatkgz

By Integration?

Like $$\int \frac{dx}{xlnx}=\int \frac{du}{u}$$

5. Oct 31, 2007

### sutupidmath

well, you could do that too, but remember u are integrating from 2 to infinity. Obviously the integral diverges, so by the Cauchy's integral theorem for the convergence of the series we can conclude that also the serie
$$\sum_{n=2}^{\infty} \frac {1}{n ln(n)}$$

diverges.

However there is another way, among many others, to show this. It is based on the comparison test and also on the mean value theorem. any way will do.

it goes like this, first we are gonna show that the series

$$\sum_{n=2}^{\infty} \{ln ln(n+1)-ln ln(n)}$$ diverges, this is pretty easy to show that diverges, because the partial sum goes to infinity as n goes to infinity.

now let us use the mean value theorem on the interval [n,n+1] for the function y=lnln(x)

lnln(n+1)-lnln(n)=(lnln(c))' where c is from the interval [n,n+1] let c=a+n, where 0<a<1, it means that c is from the interval (n,n+1), so we have

lnln(n+1)-lnln(n)=1/((a+n)ln(a+n)

so since the series $$\sum_{n=2}^{\infty} \{ln ln(n+1)-ln ln(n)}$$ diverges so will do the serie
$$\sum_{n=2}^{\infty} \frac {1}{(n+a) ln(n+a)}$$

now since

1/((a+n)ln(a+n)<1/n*ln(n) also the serie $$\sum_{n=2}^{\infty} \frac {1}{n ln(n)}$$
diverges.

i hope i was clear enough.
next time, since we are on the homework section, try to give a little more thought on your onw.

Last edited: Oct 31, 2007