# Why does $\sum_3^\infty \frac{1}{n\ln n -n}$ diverge?

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 $$\lim_{n \rightarrow \infty}\frac{\frac{1}{n\ln n-n}}{\frac{1}{n}}=\lim_{n \rightarrow \infty} \frac{1}{\ln n -1}=0$$

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

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

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

• You should give a reference or some background for the test you are using. Apr 5 at 19:43

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