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.
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 $\displaystyle p>1$
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.
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,
$\displaystyle \frac{1}{p}\ln |\ln \ln x|+C$
$\displaystyle \int \frac{1}{x\ln x (\ln \ln x)^p} dx$
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,
$\displaystyle \left\{ \begin{array}{c}\frac{(\ln \ln x)^{1-p}}{1-p}+C\mbox{ if }p\not = 1\\ \ln \ln \ln x+C \mbox{ if }p=1 \end{array} \right\}$
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 $\displaystyle p>1$