1. ## Series

For what values of p>0 does the series

a) Riemann Sum [n=1 to infinity] 1/ [n(ln n)^p]

converge and for what values does it diverge?

2. The infinite series,
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^a \ln^b n}$
Is slightly famous.
(Called Bertrand Series)

The results says that if,
$\displaystyle a>1$---> converges
$\displaystyle a<1$---> diverges
$\displaystyle a=1$---> depends.

The value of $\displaystyle b$ makes the dependance.
If $\displaystyle a=1$ then it converges only if $\displaystyle b>1$.

Therefore, your series converges when $\displaystyle p>1$

3. but i didn't learn the Bertrand Series, is there another approach to this question?

4. The function,
$\displaystyle f(x)=\frac{1}{x\ln^ p x}$
Is a positive, decreasing, continous function for $\displaystyle x>1$
With the property that,
$\displaystyle f(n)=a_n$ in the series.

You can apply the integral series test.
And see when, (I am just goint to change the lower limit, it makes no difference for convergence).
$\displaystyle \int_2^{\infty} \frac{1}{x\ln^ p x}dx$
If,
$\displaystyle u=\ln x$ then, $\displaystyle u'=1/x$
Thus, by the substitution rule,
$\displaystyle \int \frac{1}{x\ln^p x}=\int u^{-p} du$
Since $\displaystyle p>0$ the antiderivative depends whether $\displaystyle p$ is 1 or not.
So we will divide this problem into 3 cases,
$\displaystyle 0<p<1$
$\displaystyle p=1$
$\displaystyle p>1$

It, $\displaystyle 0<p<1$ then,
$\displaystyle \int u^{-p}du=\frac{u^{1-p}}{1-p}+C$
Thus, (dropping the last term, makes no difference)
$\displaystyle \int_2^{\infty} \frac{x\ln^p x}dx=\lim_{n\to\infty} \frac{\ln^{1-p} x}{1-p}$
Which will increase without bound for,
$\displaystyle 1-p>0$

If $\displaystyle p=1$ then,
$\displaystyle \int u^{-p}du=\ln u+C$
Thus, (dropping the last term, makes no difference)
$\displaystyle \int_2^{\infty} \frac{x}{\ln x}dx=\lim_{x\to\infty} \ln ln x$.
Which will increase without bound for,
$\displaystyle \ln \ln x$ is increasing (though very very very, snail slowly).

If $\displaystyle p>1$ then,
$\displaystyle \int u^{-p}du=\frac{u^{-p+1}}{-p+1}+C$
Thus, (dropping the last term, makes no difference)
$\displaystyle \int_2^{\infty} \frac{x}{\ln^p x}=\lim_{x\to \infty} \frac{\ln ^{-p+1} x}{-p+1}$
Which will converge for, $\displaystyle -p+1<0$ and this assures us the fraction is in the denominator with the denominator increasing without bound, so the limit converges.

5. another question asking for the same thing is this...

Riemann Sum [n=1 to infinity] 1/ [n(ln n) (ln(ln n))^p]

do you still use the same method? but this one seems so hard to integrate, please help. thanks.

6. Originally Posted by 413
another question asking for the same thing is this...

Riemann Sum [n=1 to infinity] 1/ [n(ln n) (ln(ln n))^p]

do you still use the same method? but this one seems so hard to integrate, please help. thanks.
The the solution is based on,
$\displaystyle \int \frac{1}{x\ln x \ln (\ln^p x)}dx$
First for $\displaystyle x\geq 3$ and $\displaystyle p>0$ (note the reason why I make these inequalities is because otherwise the function might not be defined or I will not be able to get integrals).
Then, $\displaystyle \ln (\ln^p x)=p\ln \ln x$
Thus, the integral becomes,
$\displaystyle \frac{1}{p}\int \frac{1}{x\ln x\ln \ln x}$
Express this integral as,
$\displaystyle \int \frac{1}{\ln x\ln \ln x}\cdot \frac{1}{x} dx$
If, $\displaystyle u=\ln x$ then $\displaystyle u'=1/x$
Thus,
$\displaystyle \int \frac{1}{u\ln u}\cdot u' dx$
Substitution rule,
$\displaystyle \int \frac{1}{u\ln u} du$
Express as,
$\displaystyle \int \frac{1}{\ln u}\cdot \frac{1}{u}$
Let,
$\displaystyle t=\ln u$ then, $\displaystyle t'=1/u$ thus,
$\displaystyle \int \frac{1}{t} t'du$
By the substitution rule we have,
$\displaystyle \int \frac{1}{t} dt=\ln |t|+C$
Thus,
$\displaystyle \ln |\ln u|+C$
Thus,
$\displaystyle \ln |\ln \ln x|+C$ is the integral.
Thus, the original integral is,
$\displaystyle \frac{1}{p}\ln |\ln \ln x|+C$

7. but the exponent p is outside the second ln, does that make a difference?

1/ [n(ln n) (ln(ln n))^p]

8. is the answer divergent for all p?

9. Originally Posted by 413
but the exponent p is outside the second ln, does that make a difference?

1/ [n(ln n) (ln(ln n))^p]
$\displaystyle \int \frac{1}{x\ln x (\ln \ln x)^p} dx$

Let,
$\displaystyle u=\ln x$ then, $\displaystyle u'=1/x$
Thus,
$\displaystyle \int \frac{1}{u (\ln u)^p}du$
Let,
$\displaystyle t=\ln u$ then, $\displaystyle t'=1/u$
Thus,
$\displaystyle \int \frac{1}{t^p}dt=\int t^{-p}dt=\left\{ \begin{array}{c}\frac{t^{1-p}}{1-p}+C\mbox{ if }p\not = 1\\ \ln t +C\mbox{ if }p=1 \end{array} \right\}$
That is how you integrate it.

Double substitution,
$\displaystyle \left\{ \begin{array}{c}\frac{(\ln \ln x)^{1-p}}{1-p}+C\mbox{ if }p\not = 1\\ \ln \ln \ln x+C \mbox{ if }p=1 \end{array} \right\}$

10. it is still divergent for all p right?

11. Originally Posted by 413
it is still divergent for all p right?
How! Did you try to solve it? No.

Look at the integral. For the same reason it will diverge for $\displaystyle p=1$ for $\displaystyle \ln \ln \ln x$ is divergent.

And if the exponent $\displaystyle 1-p>0$ then you have again an increasing function so it is divergent. When $\displaystyle 1-p<0$ then the exponent is negative thus the function will eventually die. In that case is convergese for $\displaystyle p>1$