Series

**413** · November 11th 2006, 06:39 PM

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?

I am having trouble doing this question, please help.

**ThePerfectHacker** · November 11th 2006, 07:14 PM

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

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

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

Therefore, your series converges when $p>1$

**413** · November 12th 2006, 07:23 AM

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

**ThePerfectHacker** · November 12th 2006, 08:18 AM

The function,
$f(x)=\frac{1}{x\ln^ p x}$
Is a positive, decreasing, continous function for $x>1$
With the property that,
$f(n)=a_n$ 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).
$\int_2^{\infty} \frac{1}{x\ln^ p x}dx$
If,
$u=\ln x$ then, $u'=1/x$
Thus, by the substitution rule,
$\int \frac{1}{x\ln^p x}=\int u^{-p} du$
Since $p>0$ the antiderivative depends whether $p$ is 1 or not.
So we will divide this problem into 3 cases,
$0<p<1$
$p=1$
$p>1$

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

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

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

**413** · November 12th 2006, 11:35 AM

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.

**ThePerfectHacker** · November 12th 2006, 03:38 PM

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,
$\int \frac{1}{x\ln x \ln (\ln^p x)}dx$
First for $x\geq 3$ and $p>0$ (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, $\ln (\ln^p x)=p\ln \ln x$
Thus, the integral becomes,
$\frac{1}{p}\int \frac{1}{x\ln x\ln \ln x}$
Express this integral as,
$\int \frac{1}{\ln x\ln \ln x}\cdot \frac{1}{x} dx$
If, $u=\ln x$ then $u'=1/x$
Thus,
$\int \frac{1}{u\ln u}\cdot u' dx$
Substitution rule,
$\int \frac{1}{u\ln u} du$
Express as,
$\int \frac{1}{\ln u}\cdot \frac{1}{u}$
Let,
$t=\ln u$ then, $t'=1/u$ thus,
$\int \frac{1}{t} t'du$
By the substitution rule we have,
$\int \frac{1}{t} dt=\ln |t|+C$
Thus,
$\ln |\ln u|+C$
Thus,
$\ln |\ln \ln x|+C$ is the integral.
Thus, the original integral is,
$\frac{1}{p}\ln |\ln \ln x|+C$

**413** · November 12th 2006, 03:53 PM

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

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

**413** · November 14th 2006, 08:41 AM

is the answer divergent for all p?

**ThePerfectHacker** · November 14th 2006, 09:13 AM

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]

$\int \frac{1}{x\ln x (\ln \ln x)^p} dx$

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

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

**413** · November 14th 2006, 09:17 AM

it is still divergent for all p right?

**ThePerfectHacker** · November 14th 2006, 09:23 AM

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 $p=1$ for $\ln \ln \ln x$ is divergent.

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

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