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

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.

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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. – Henning Makholm Feb 13 '12 at 1:09
I'm guessing the lower limit is $n=1$ – Pedro Tamaroff Feb 13 '12 at 1:30
@Peter Yes. I have closed this. Thank you. – MathMathCookie Feb 13 '12 at 1:36
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. – Aryabhata Feb 13 '12 at 2:57

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