1
$\begingroup$

Does the series $\sum_{n=2}^{\infty} (-1)^{n+1}\frac{\ln (n)}{n}$ converge? If so, does it converge absolutely?

My attempt:

Because $$\left(\frac{\ln x}{x}\right)' = \frac{1-\ln(x)}{x²} > 0$$ whenever $x > e$, and because $\ln(n)/n \to 0$, the Leibniz criterion tells us that the series converges.

Because $$\int_{1}^{+ \infty} \frac{\ln(x)}{x}dx = \lim_{k \to \infty} 1/2\ln^2(k) = + \infty$$

the series does not converge absolutely. Is this correct?

$\endgroup$
  • $\begingroup$ Geez, seems like a lot of convergence questions from you all at once. Slow down! $\endgroup$ – Frpzzd Dec 25 '17 at 22:49
  • $\begingroup$ If I would post them in one post, people will tell me to post them separately. What do you suggest I do as I don't have a teacher to give my solutions? $\endgroup$ – user370967 Dec 25 '17 at 22:50
  • $\begingroup$ Correct, that shows the series does not converge absolutely. Check next if it fits the criteria of convergence in the alternating series test. $\endgroup$ – KCd Dec 25 '17 at 22:57
  • 2
    $\begingroup$ If you do not want to put everything in one post, you can also exercise patience and not post everything at the same time in different posts. As you learn more, you should develop the capacity to know yourself when you have solved a problem correctly. $\endgroup$ – KCd Dec 25 '17 at 22:58
  • $\begingroup$ @Math_QED There's no need to check all of your answers all at once. After checking a few of your answers and getting them all right, you should be sure enough of yourself to not have to check all of them. Also, if you must, WolframAlpha should be able to check convergence for you. Where do these problems come from? $\endgroup$ – Frpzzd Dec 25 '17 at 22:58
1
$\begingroup$

More simply, without calculating the integral, it is because $\dfrac{\ln n}n>\dfrac 1n$ if $n>2$, and the harmonic series diverges.

$\endgroup$
0
$\begingroup$

You can show with $\displaystyle S_{2n}=\sum_{k=1}^n\bigg(\dfrac{\ln(2k+1)}{2k+1}-\dfrac{\ln2k}{2k}\bigg)$ that $\bigg(\dfrac{\ln(2n+1)}{2n+1}-\dfrac{\ln2n}{2n}\bigg)=o(n^{-3/2})$ so $S_{2n}$ converges.

$\endgroup$
  • $\begingroup$ This needs $x > \ln(1+x) > x-x^2/2$ or something equivalent. This can be proved from the integral definition of $\ln$. $\endgroup$ – marty cohen Dec 26 '17 at 7:14

Your Answer

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