$$\sum_{n=2}^{\infty} \frac{1}{(n\log(n))^p}.$$ Let $$a_{n}:= \frac{1}{(n\log(n))^p}$$

for $p \leq 0 $ divergent by $n^{th}$ term test.

for $p>0$ $a_n$ is non negative decreasing

We'll use the Cauchy Condensation test

$$\sum_{n=2}^{\infty} \frac{2^n}{(2^n\log(2^n))^p} = \sum_{n=2}^{\infty} \frac{2^n}{2^{np}n^p (\log(2))^p} = \frac{1}{(\log(2))^p}\sum_{n=2}^{\infty} \frac{2^n}{2^{np}n^p}$$

$$ (\frac{2^n}{2^{np}n^p})^{1/n} = \frac{1}{2^{p-1}n^{p/n}} \to \frac{1}{2^{p-1}}$$

For convergence $$\frac{1}{2^{p-1}} < 1 \Rightarrow p-1>0 \Rightarrow p >1 $$

and divergent for $0<p<1$

for $p=1$

$$\sum_{n=2}^{\infty} \frac{1}{n\log(n)} $$ diverges by integral test.

Conclusion : Converges if $p>1$ otherwise diverges

Is this argument fine?

  • $\begingroup$ Just a note, please use \log as opposed to log, makes the formatting nicer. (Compare $\log$ with $log$). $\endgroup$ – Rhys Hughes Jun 5 '18 at 16:38
  • $\begingroup$ noted, thanks:) $\endgroup$ – mathlad Jun 5 '18 at 16:41
  • $\begingroup$ You can do this entirely by comparison. For $0<p<1$ compare to $n^{-(p+\epsilon)}.$ For $p>1,$ compare to $n^{-p}.$ (And $p=1$ can be done by the integral test like you say.) $\endgroup$ – spaceisdarkgreen Jun 5 '18 at 22:52

I'm a bit rusty on the Cauchy Condensation test. But here is an alternative solution, perhaps this will be of some use.

Let $a_k=\bigg( \frac{1}{k\ln k} \bigg)^p$

First, if $p\leqq 0 $ clearly $a_k \not\rightarrow 0$ as $k\rightarrow \infty$ and thus $\sum_{k=2}^{\infty} a_k$ diverges.

Now choose $b_k=\frac{1}{k(\ln k)^p}$ such that $\sum_{k=2}^{\infty} b_k$ is convergent for $p>1$ and divergent for $p \in (0,1]$. This follows from the integral test since $b_k$ is positive and decreasing $\forall k\in[2,\infty]$.

Note that in general if $|\frac{a_k}{b_k}| \longrightarrow 0, k\longrightarrow \infty$ it means that $\forall \epsilon>0$ $ \exists$ $M>0$ such that $|a_k| < \epsilon|b_k|$ when $k>M$ and if $\sum a_k$ and $\sum b_k$ are both positive series, we therefore know that $\sum b_k$ convergent $\implies$ $\sum a_k$ convergent by the comparison test.

In the same manner, we have that if $|\frac{a_k}{b_k}| \longrightarrow \infty, k\longrightarrow \infty$ $\implies$ $\forall N>0$ $\exists$ $M>0$ such that $|\frac{a_k}{b_k}| > N $ if $k>M$ and therefore $\sum b_k$ divergent $\implies \sum a_k$ diverges.

In our case, fix $p\in(0,1)$, then $|\frac{a_k}{b_k}| = \frac{1}{k^{p-1}} \longrightarrow \infty, k \longrightarrow \infty$ and because we know that $\sum b_k$ diveriges for $p\in(0,1)$ so does $\sum a_k$ (since both $a_k,b_k$ are non negative). If $p=1$ we use the integral test as above.

Now fix $p>1$ we have $|\frac{a_k}{b_k}| = \frac{1}{k^{p-1}} \longrightarrow 0, k \longrightarrow \infty$ and since $\sum b_k$ converges for these $p$, so does $\sum a_k$.

Thus, $\sum_{k=2}^{\infty}\bigg( \frac{1}{k\ln k} \bigg)^p$ converges $\forall$ $p>1$.


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