I am trying to prove that the series $$\sum_1^\infty \frac{1}{(k+1)\ln^\alpha(k+1)}$$ converges for $\alpha >1$ and diverges for $\alpha <1$

I understand the proof using the integral test, but we haven't built the integral yet. We do have the p-test, Cauchy condensation, and the ratio test. I'm not totally sure where to start without the integral test.

Intuitively, using the comparison and p-tests, $\lim_{k \to \infty} \frac{1}{(k+1)\ln^\alpha(k+1)}$ approaches $0$ more quickly than $\lim_{k \to \infty} \frac{1}{k^p}$ with p = 1 if $\ln^\alpha(k+1) > 1$, but that seems to indicate convergence for $\alpha > 0$, rather than $1$

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    $\begingroup$ Cauchy condensation is specially designed for this one... $\endgroup$ – zwim Dec 3 '17 at 4:40
  • $\begingroup$ My edit was extra backets, as $ \ln (k+1)^a$ could be read as $\ln ((k+1)^a)$ $\endgroup$ – DanielWainfleet Dec 3 '17 at 7:49

Exercise: Let $D\subset \Bbb R$. Let $f:D\to [0,\infty)$ and $g:D\to [0,\infty)$ be increasing (i.e. for $a, b\in D,$ if $a<b$ then $f(a)\leq f(b)$ and $g(a)\leq g(b)$. And let $a>0.$ then (I)...$h(x)=f(x)g(x)$ is increasing, and (II)...$j(x)=g(x)^a$ is increasing.

Applying (II) with $D=[1,\infty)$ and $g(x)=\ln x,$ the function $j(x)=(\ln x)^a$ is increasing on $[1,\infty)$ when $a>0$. Applying (I) with $D=[1,\infty)$ and $f(x)=x$ and $g(x)=(\ln x)^a,$ with $a>0 ,$ the function $h(x)=x(\ln x)^a$ is increasing on $[1,\infty).$

So apply the Cauchy Condensation test. With $a(n)=[ n(\ln n )^a]^{-1}$ for $a>1,$ we have $2^na(2^n)=2^n[ 2^n\cdot (n\ln 2)^a)]^{-1}= [n\ln 2]^{-a}=b(n) $ which is decreasing if $a>0.$

So apply the Cauchy test again, to $b(n)=[n\ln 2]^{-a}.$ We have $2^n b(2^n)=2^n[2^n\ln 2]^{-a} =(\ln 2)^{-a} \cdot (2^{a-1})^n=c(n).$

Now $\sum_n c(n)$ is a geometric series. It converges iff $2^{a-1}>1$ iff $a>1.$

  • $\begingroup$ Thanks, this makes perfect sense! $\endgroup$ – John Doe Dec 3 '17 at 13:31
  • $\begingroup$ Thank you. Note that the Cauchy Condensation Test for convergence of $\sum_n x_n$ works if $x_n$ is monotonic for all sufficiently large n, not necessarily for all $n$.... You may want to search this site for answered Q's about various convergent tests & theorems. $\endgroup$ – DanielWainfleet Dec 3 '17 at 18:57
  • $\begingroup$ Awesome, thanks, I am doing that as I go, any suggestions for good resources/practice sets beyond just variations of like "convergent(ce) test for series" or "prove series converges/diverges" $\endgroup$ – John Doe Dec 3 '17 at 22:50
  • $\begingroup$ The classic old book "Infinite Sequences And Series" by Bromwich, which requires almost no pre-requisites. $\endgroup$ – DanielWainfleet Dec 3 '17 at 23:34

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