2
$\begingroup$

Evaluate if the following series is convergent or divergent: $\sum\limits_{n=2}^\infty \frac {1} {\ln{n}}$.

I solved the problem using Weierstrass's comparison theorem once the integral would get messy:

$n+c\geqslant \ln(n)$, where $c$ is a constant. We can check that $$\frac{d\ln(x)}{dx}=\frac{1}{x}<1=\frac{dy}{dx},$$ for $y=x$ and $x\in\mathbb{N}$, which shows that $\ln(n)$ grows slower than $n$.

Then $$\sum\limits_{n=2}^\infty \frac {1} {{n}} <\sum\limits_{n=2}^\infty \frac {1} {{n}+c}<\sum\limits_{n=2}^\infty \frac {1} {\ln{n}},$$ proving the series diverge.

Question:

Is my answer right? If not why? What are other possibilities?

Thanks in advance!

$\endgroup$
  • $\begingroup$ What "Weierstrass Comparison Theorem"?? $\endgroup$ – DonAntonio Apr 1 '18 at 11:01
  • $\begingroup$ @DonAntonio It is a proof of the comparasion test. People call it Weierstrass's test in honour of the German mathematician that carried that surname. $\endgroup$ – Pedro Gomes Apr 1 '18 at 11:04
  • $\begingroup$ The only Weierstrass Test I know is Weierstrass's $\;M\,-$ test, for power series. I'm not sure what you call Weierstrass Comparison theorem...perahsp you mean the usual comparison test for positive series? $\endgroup$ – DonAntonio Apr 1 '18 at 11:07
  • $\begingroup$ @DonAntonio That is precisely what I mean. $\endgroup$ – Pedro Gomes Apr 1 '18 at 11:10
  • 1
    $\begingroup$ Another alternative: $\frac{1}{\ln n} > \frac{1}{n \ln n}$ and the latter series diverges by the integral test $\int \frac{dx}{x\ln x} = \ln\ln(x) \to \infty$. $\endgroup$ – Winther Apr 1 '18 at 11:24
4
$\begingroup$

You can simply say that $(\forall n\in\mathbb{N}):\ln n<n$ and that therefore$$\sum_{n=2}^\infty\frac1{\ln n}\geqslant\sum_{n=2}^\infty\frac1n=+\infty.$$

$\endgroup$
2
$\begingroup$

Use that $$n>\log(n)$$ for $n>0$

$\endgroup$
2
$\begingroup$

Use the fact that $$n>\log(n)$$ So that we get $$\sum_{n=2}^\infty\frac{1}{\ln n}> \sum_{n=2}^\infty\frac{1}{n}=\infty$$

$\endgroup$
1
$\begingroup$

Use Cauchy's condensation test with $2^n > n$. (Easily proved by induction/binomial theorem.)

$$2^n a_{2^n} = \frac{2^n}{\ln 2^n} = \frac{2^n}{n \ln 2} > \frac{1}{\ln2}$$

So the series diverges.

$\endgroup$

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.