1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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 $\forall$ $p>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.