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Let $p,q\in\mathbb{R}$ and consider the series $\sum_{n\geq 2}\frac{1}{n^{p}(\ln(n))^{q}}$.

i) Show by the comparison test, that the series is convergent if $p>1$ and divergent if $p<1$ for all $q\in\mathbb{R}$.

ii) Show by the integral test, that if $p=1$ the series is convergent if and only if $q>1$.

[Hint: The antiderivatives $(\ln(x))^{-a}$ and $\ln(\ln(x))$ can be used].

My answer to i) Define $a_{n}:=\frac{1}{n^{p}(\ln(n))^{q}}$. Note that $\ln(x)<1/x\implies 1/\ln(x)>x$ for all $x\in [2,\infty)$. We can use the comparisons:

If $q\geq 0$ we have $1/(\ln(x))^{q}\geq 1/x^{q}$ and $1/(\ln(x))^{q}\leq x^{q}$.

If $q< 0$ we have $1/(\ln(x))^{q}<1/x^{q}$ and $1/(\ln(x))^{q}>x^{q}$.

Assume $q\geq 0$. We see that $\sum_{n\geq 2}a_{n}\geq \sum_{n\geq2}1/x^{p+q}$. By divergence test, the $\sum_{n\geq2}1/x^{p+q}$ diverges if $p+q<1 \iff p<1-q\leq 1$, that is, if $p\leq 1$ (or should it be $p<1$?)

We also see that $\sum_{n\geq 2}a_{n}\leq \sum_{n\geq2}1/x^{p-q}$. By the $p$-test, the $\sum_{n\geq2}1/x^{p-q}$ converges if $p-q>1\iff p>1+q\geq 1$, that is, if $p>1$.

Assume $q<0$. We see that $\sum_{n\geq 2}a_{n}<\sum_{n\geq 2}1/n^{p}$ which converges if $p>1$. We also see that $\sum_{n\geq 2}a_{n}>\sum_{n\geq 2}1/n^{p-q}$ which diverges if $p-q<1\iff p<1+q<1$.

Question: I have shown that the series diverges if $p\leq 1$ for $q\geq 0$ -- but it diverges if $p<1$ for $q<0$. What do I have to do with it?

My answer to ii) I don't know how the hint is helping me. Define $f(x):=\frac{1}{x^{p}(\ln(x))^{q}}$ for all $x\geq 2$. The function is decreasing if $f'(x)\leq 0 \iff q\geq -p\ln(x)$, that is, if $x\geq e^{-q/p}$. Letting $p=1$, the function is decreasing if $x\geq e^{-q}$.

$\impliedby:$ For $q\neq 1$, we have that $$ \int_{2}^{\infty}f(x)\,\mathrm{d}x=\int_{2}^{\infty}\frac{1}{x(\ln(x))^{q}}=\int_{\ln(2)}^{\infty}\frac{1}{m^{q}}\,\mathrm{d}m=\left [ \frac{1}{1-q}\frac{1}{m^{q-1}} \right ]_{\ln(2)}^{\infty}. $$ We see that the series is convergent if $q-1>0$ and divergent if $q-1<0$. For $q=1$, the series too is divergent.

$\implies$: Proving it by contraposition that the series is divergent if $q\leq 1$. Since it already has been shown, then the series converges only if $q>1$.

Edit: I have edited both cases many times. Thanks to Steven Stadnicki, Dr. MV and Bernard for helping me.

Mittens
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UnknownW
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  • For case (ii) the hypothesis is $p=1,$! – Bernard May 03 '15 at 22:10
  • @Bernard. Do you mean, that (ii) says letting $p = 1$? So that I have to show that the series converges if and only if $p>1$, that is, I have to show that it converges if $q>1$ and it diverges if $q\leq 1$ by contraposition? – UnknownW May 03 '15 at 23:14
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    I mean that, yes. But the series converges if (and only if) $p>1$ or ($p=1,$ and $,q>1$). This kind of series is known as a Bertrand's series. – Bernard May 03 '15 at 23:22
  • Thank you for informing. I've changed completely in my answer to the second case. I hope it is correct now. Could you please help me with the first case? – UnknownW May 04 '15 at 16:39
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    The statement that $\ln(n)^q\geq n^q$ if $q\lt 1$ isn't true; you need $q\lt 0$ for the inequality to reverse. If, for instance, $q=1/2$ then $\sqrt{\ln(n)}\lt\sqrt{n}$ as $n\to\infty$. Perhaps you just mean $q\lt 0$ there? – Steven Stadnicki May 04 '15 at 16:40
  • @StevenStadnicki. Well, I totally forgot that $q$ is defined to be a real number, not a integer number. I would then mean $q\leq 0$. But I have to find a comparison for $q<1$. Can you suggest some? – UnknownW May 04 '15 at 16:55
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    @AjmalW For $0\lt q\lt 1$ your original comparison is just fine for showing convergence; you still have $(\ln n)^q\gt 1$ so $\frac1{n^p(\ln n)^q}\lt\frac1{n^p}$. For $q\lt 0$ you need to be a little trickier. – Steven Stadnicki May 04 '15 at 17:51
  • @StevenStadnicki I have corrected my answer to i). Do you have time to look at it, please? – UnknownW May 04 '15 at 20:22

2 Answers2

1

Let $I$ denote the integral

$$I=\int_2^{L}\frac{1}{x(\log x)^q}dx=\frac{1}{1-q}\left((\log L)^{1-q}-(\log 2)^{1-q}\right)$$

for $q\ne 1$. For $q=1$, $I=\log \log (L/2)$

Notice that if $q>1$, then $\lim_{L\to \infty}I=-(\log 2)^{1-q}/(1-q)$ while if $q\le 1$, the integral diverges.

Thus, by the integral test, the series converges for $q>1$ and diverges otherwise as expected!

NOTE

The integration can be performed by making the substitution $u=\log x$ so that $du=dx/x$. Then, $I=\int_{\log 2}^{\log L} \frac{1}{u^q}du$.

Gary
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Mark Viola
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0

Here is another for solution for $p=q$ based on Kummer-Bertrand's test.

  • Suppose $p\leq 1$. For all $n\geq 2$ $\frac{n\log n}{(n+1)\log(n+1)}\leq 1$; hence $$n\log n-(n+1)\log(n+1)\frac{n^p\log^p(n)}{(n+1)^p\log^p(n+1)}=n\log n\Big(1-\frac{n^{p-1}\log^{p-1}(n)}{(n+1)^{p-1}\log^{p-1}(n+1)}\Big)\leq 0$$ Hence, by Kummar-Betrand's theorem, $\sum_n\frac{1}{n^p\log^pn}$ diverges.

  • Suppose $p>1$. \begin{align} \frac{n\log n}{(n+1)\log(n+1)}&=\frac{n \log (n+1)+ \log(1-\frac{1}{1+n})^n}{(n+1)\log(n+1)}\\ &=1-\frac{1}{n+1}+ \frac{\log(1-\frac{1}{1+n})^n}{(n+1)\log(n+1)}\\ &=1-\frac{1}{n+1}+ r_n \end{align} where $(n+1)\log(n+1)r_n\xrightarrow{n\rightarrow\infty}-1$. Then $$\Big(\frac{n\log n}{(n+1)\log(n+1)}\Big)^{p-1}=1-(p-1)\big(\frac{1}{n+1}-r_n\big)+o\Big(\frac{1}{n+1}-r_n\Big)$$ Hence $$n\log n\Big(1-\frac{n^{p-1}\log^{p-1}(n)}{(n+1)^{p-1}\log^{p-1}(n+1)}\Big)=(p-1)\log n\frac{n}{n+1}-(p-1)r_n n\log n+n\log n o\Big(\frac{1}{n+1}-r_n\Big) $$ whence we obtain that $$\lim_n n\log n\Big(1-\frac{n^{p-1}\log^{p-1}(n)}{(n+1)^{p-1}\log^{p-1}(n+1)}\Big)=\infty$$ By Kummer-Bertrnad's theorem, the series $\sum_n\frac{1}{n^p\log^p(n)}$ converges.

Mittens
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