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I was not able to find a function to use direct comparison with this series. $\frac{1}{n}$ and $\frac{1}{\ln(n)}$ are both larger than $\frac{1}{n\ln n-n}$.

Using the limit comparison test does not help since $$\lim_{n \rightarrow \infty}\frac{\frac{1}{n\ln n-n}}{\frac{1}{n}}=\lim_{n \rightarrow \infty} \frac{1}{\ln n -1}=0$$

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    $\begingroup$ Can you compare to an integral? $\endgroup$
    – J.G.
    Apr 5 at 19:02
  • $\begingroup$ I see. $ \frac{1}{n \log n} $ is smaller than $ \frac{1}{n \log n -n} $ $\endgroup$
    – Will Jagy
    Apr 5 at 19:12
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    $\begingroup$ $\frac{1}{n\ln(n)-n}>\frac{1}{n\ln(n)}=a_n$. Prove that $a_n$ is decreasing and apply Cauchy’s condensation test. $\endgroup$
    – Antonio
    Apr 5 at 19:18
  • $\begingroup$ @J.G. doing the integral scared me at first but I realized it's just a simple u-sub problem lol $\endgroup$ Apr 5 at 19:23
  • $\begingroup$ Also, you need only compare to $\int\frac{dx}{x\ln x}$ rather than the harder $\int\frac{dx}{x\ln x-x}$. $\endgroup$
    – J.G.
    Apr 5 at 19:31

2 Answers 2

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As the sequence $a_n=n\ln(n)-n=n(\ln(n)-1)$ is monotone and $a_n>0$, we can use the following test: $\sum a_n$ converges if and only if the sum $\sum 2^na_{2^n}$ converges. We get

$b_n=2^na_{2^n}=2^n\dfrac{1}{2^n(\ln(2^n)-1)}=\dfrac 1{\ln(2)n-1}$. This sum diverges.

The test follows from the following inequality

$2(a_{2^{n-1}+1}+\dots+ a_{2^n}) \leq 2^na_{2^n}\leq a_{2^n+1}+\dots+a_{2^{n+1}} $.

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    $\begingroup$ You should give a reference or some background for the test you are using. $\endgroup$
    – Rob Arthan
    Apr 5 at 19:43
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Just do the integral test

$$ \int_{2}^\infty \frac{dx}{x\ln (x)-x}= \int_{2}^\infty \frac{dx}{x(\ln (x)-1)}= \int_{\ln(2)}^{\infty} \frac{du}{u-1} = \ln (+\infty-1) -\ln(\ln(2)-1) \to +\infty$$

so the series diverges

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