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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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    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.2012-02-13

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