1
$\begingroup$

Does the infinite sum $\displaystyle \sum\frac{(n+1)}{\ln(n+1)}$ converge?

I actually know it doesn't since if we use the integral test, and let $\ln(n+1)=u$ and $\displaystyle du=\frac{1}{n+1}dx$, then we have $$\int\frac{du}{u}\;,$$ which gives us $\ln(u)=\ln(\ln(n+1))$, which diverges. But, is there another way to find this?

I apologize for the format, not knowing Latex, I used WolframAlpha before to try to get the syntax, but they seemed to have made that a Pro account only feature now. I will make an effort to learn it in the future.

$\endgroup$
4
  • 10
    $\begingroup$ Do you mean $\sum_{n=0}^\infty \frac{n+1}{\log(n+1)}$? Then the terms don't even go towards zero, so clearly the sum doesn't converge. $\endgroup$
    – hmakholm left over Monica
    Feb 13, 2012 at 1:09
  • $\begingroup$ I'm guessing the lower limit is $n=1$ $\endgroup$
    – Pedro ♦
    Feb 13, 2012 at 1:30
  • $\begingroup$ @Peter Yes. I have closed this. Thank you. $\endgroup$
    – MathMathCookie
    Feb 13, 2012 at 1:36
  • 2
    $\begingroup$ And if the term was inverted, $\sum_{k=0}^{n} \frac{\log (k+1)}{k+1} \ge C + \sum_{k=2}^{n} \frac{1}{k+1}$ and hence does not converge either. $\endgroup$
    – Aryabhata
    Feb 13, 2012 at 2:57

1 Answer 1

Reset to default
3
$\begingroup$

$\lim\limits_{x\rightarrow \infty} \frac{x}{\ln x}$ does not exist. So the series does not converge.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .