# Thread: [SOLVED] Infinite series, convergent or not?

1. ## [SOLVED] Infinite series, convergent or not?

It has probably been answered before on MHF, but I didn't find it using the search button.
Tell whether the following infinite series is convergent or not :
$\displaystyle \sum_{n=2}^{+\infty} \frac{1}{n \ln (n)}$. At first, I used the ration test, but I cannot conclude since it says "1". Then the integral test, which gave me divergent, but I'm not sure I did it well, because when I tried to get an approximation via Mathematica, I got an error like if it converges, it does it too slow or something like that. Nevertheless, it said an approximation is 13.4391. From it I doubted my integral test was well made. I tried the root test, but I'm not sure how to finish it.
Integral test : $\displaystyle \int_2^{+\infty} \frac{1}{x\ln (x)}$. Let $\displaystyle u$ be $\displaystyle \ln (x)$, then $\displaystyle du=\frac{1}{x}$, and the integral becomes $\displaystyle \int_2^{+\infty} \frac{du}{u}=\ln u \big | ^{+\infty}_2$. Replacing $\displaystyle u$ by $\displaystyle \ln (u)$, it's easy to see the integral diverges.
So normally, so does the infinite series. Am I right?

2. Originally Posted by arbolis
It has probably been answered before on MHF, but I didn't find it using the search button.
Tell whether the following infinite series is convergent or not :
$\displaystyle \sum_{n=2}^{+\infty} \frac{1}{n \ln (n)}$. At first, I used the ration test, but I cannot conclude since it says "1". Then the integral test, which gave me divergent, but I'm not sure I did it well, because when I tried to get an approximation via Mathematica, I got an error like if it converges, it does it too slow or something like that. Nevertheless, it said an approximation is 13.4391. From it I doubted my integral test was well made. I tried the root test, but I'm not sure how to finish it.
Integral test : $\displaystyle \int_2^{+\infty} \frac{1}{x\ln (x)}$. Let $\displaystyle u$ be $\displaystyle \ln (x)$, then $\displaystyle du=\frac{1}{x}$, and the integral becomes $\displaystyle \int_2^{+\infty} \frac{du}{u}=\ln u \big | ^{+\infty}_2$. Replacing $\displaystyle u$ by $\displaystyle \ln (u)$, it's easy to see the integral diverges.
So normally, so does the infinite series. Am I right?
in terms of u, the limits are $\displaystyle \int_{\ln 2}^{\infty} \frac{du}{u}=\ln u \bigg | ^{\infty}_{\ln 2}$

yes, it is divergent

3. Originally Posted by arbolis
It has probably been answered before on MHF, but I didn't find it using the search button.
Tell whether the following infinite series is convergent or not :
$\displaystyle \sum_{n=2}^{+\infty} \frac{1}{n \ln (n)}$. At first, I used the ration test, but I cannot conclude since it says "1". Then the integral test, which gave me divergent, but I'm not sure I did it well, because when I tried to get an approximation via Mathematica, I got an error like if it converges, it does it too slow or something like that. Nevertheless, it said an approximation is 13.4391. From it I doubted my integral test was well made. I tried the root test, but I'm not sure how to finish it.
Integral test : $\displaystyle \int_2^{+\infty} \frac{1}{x\ln (x)}$. Let $\displaystyle u$ be $\displaystyle \ln (x)$, then $\displaystyle du=\frac{1}{x}$, and the integral becomes $\displaystyle \int_2^{+\infty} \frac{du}{u}=\ln u \big | ^{+\infty}_2$. Replacing $\displaystyle u$ by $\displaystyle \ln (u)$, it's easy to see the integral diverges.
So normally, so does the infinite series. Am I right?

$\displaystyle \int_2^{\infty}\frac{dx}{x\ln^p(x)}$

Now lets look at the five cases

Case one

If $\displaystyle p<0$

$\displaystyle \int_2^{\infty}\frac{dx}{x\ln^p(x)}$ can be rewritten as $\displaystyle \int_2^{\infty}\frac{\ln^{-p}(x)}{x}dx$ where$\displaystyle -p>0\quad\text{because }p<0$

So then

$\displaystyle \int_2^{\infty}\frac{\ln^{-p}(x)}{x}dx>\int_2^{\infty}\frac{dx}{x}$

Therefore divergent.

Now what about when $\displaystyle p=0$

Obviously then we have

$\displaystyle \int\frac{dx}{x\ln^0(x)}=\int_1^{\infty}\frac{dx}{ x}=\infty$

So divergent.

$\displaystyle 0<p<1$

We then have that

$\displaystyle \int\frac{dx}{x\ln^p(x)}dx=\frac{\ln^{-p+1}(x)}{-p+1}$

Now at $\displaystyle x=2\Rightarrow\frac{\ln^{-p+1}(2)}{-p+1}\text{ Which is a finite value}$

Now for $\displaystyle x\to\infty$ consider that since $\displaystyle 0<p<1\Rightarrow{-p+1>0}$

And it can can be shown by direct substitution that $\displaystyle \lim_{x\to\infty}\ln^{n\in(0,1)}(x)=\infty$

Therefore divergent.

Now when $\displaystyle p=1$ we have that

$\displaystyle \int_2^{\infty}\frac{dx}{x\ln(x)}=\ln(\ln(x))\bigg |_2^{\infty}=\infty-\ln(\ln(2))$

So divergent.

$\displaystyle p>1$

we have that

$\displaystyle \int_2^{\infty}\frac{dx}{x\ln^p}=\frac{\ln^{-p+1}(x)}{-p+1}\bigg|_{2}^{\infty}$

Now consider that

$\displaystyle p>1\Rightarrow{-p+1<0}$

Now it can be shown very easily that $\displaystyle \lim_{x\to\infty}\ln^{n\in(-\infty,0)}(x)=0$

So then

$\displaystyle \int_2^{\infty}\frac{dx}{x\ln^p(x)}=0-\frac{\ln^{-p+1}(2)}{-p+1}$

Therefore convergent.

Putting this together we get

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

Converges only for $\displaystyle p>1$