Convergence of $\sum \frac{\ln n}{n^2 -3n}$

Question

Does $\sum_{n=4} \frac{\ln n}{n^2 -3n}$ converge? My "gut feeling" tells me that it converges since $n^2$ grows faster than $\ln(n)$. But I cannot come up with an argument. I feel that comparison test is the way to go. I tried

$$\sum \frac{\ln n}{n^2 - 3n}<\sum\frac{n}{n^2 - 3n}=\sum\frac{1}{n-3},$$

which doesn't work. I also tried

$$\sum\frac{\ln n}{n^2 - 3n}<\sum\frac{\sqrt{n}}{n^2 -3n}<\sum\frac{1}{n^\frac{3}{2}-3\sqrt{n}},$$ since I want a p-series that converges. I want to get rid of the $3\sqrt{n}$, but not sure how. Specifically, we do not have

$$\sum\frac{1}{n^\frac{3}{2}-3\sqrt{n}}<\sum\frac{1}{n^\frac{3}{2}}$$

Answer 1

Actually, your second one does show convergence. The $3\sqrt{n}$ can be ignored, and $\sum_{n=1}^{\infty} \frac1{n^p} $ converges for $p > 1$.

To show why the $3\sqrt{n}$ can be ignored, note that

$\begin{array}\\ \dfrac1{n^{3/2}-3\sqrt{n}} &=2\dfrac1{2n^{3/2}-2\sqrt{n}}\\ &=2\dfrac1{n^{3/2}+n^{3/2}-2\sqrt{n}}\\ &\le 2\dfrac1{n^{3/2}}\\ \end{array} $

when $n^{3/2} > 2\sqrt{n}$ or $n > 2$.

Answer 2

Use a comparison test with, say, $\sum_n n^{-3/2}$.

$$ \dfrac{\ln(n)/(n^2-3n)}{n^{-3/2}} = \dfrac{\ln n}{n^{1/2} (1 - 3/n)} \to 0\ \text{as}\ n \to \infty$$

Answer 3

For every $\delta>0 $ we have that exists some $n_{0} $ such that $\log\left(n\right)<n^{\delta},\,\forall n\geq n_{0} $. So $$\sum_{n\geq n_{0}}\frac{\log\left(n\right)}{n^{2}-3n}<\sum_{n\geq n_{0}}\frac{1}{n^{2-\delta}-3n^{1-\delta}}\sim\sum_{n\geq n_{0}}\frac{1}{n^{2-\delta}}$$ so it is sufficient to take $\delta<1$ and we have done.

Answer 4

The general term is equivalent to $$\frac{\ln n}{n^2-3n}\sim_\infty\frac{\ln n}{n^2}=\frac1{n^2\ln^{-1}n},$$ which is a convergent Bertrand's series.

Reminder: A Bertrand's series: $\;\displaystyle\sum_{n\ge2}\frac1{n^{\alpha}\ln^{\beta}n}$, is convergent if $\;\alpha>1\;$ or $\;\alpha=1, \,\beta>1$, divergent otherwise.

Answer 5

You could show the convergence using standard tests.

Let us try the ratio test $$u_n=\frac{\log (n)}{n^2 -3n}$$ $$\frac{u_{n+1}}{u_n}=\frac{\log(n+1)}{\log(n)}\times\frac{n^2 -3n}{n^2-n-2}=\frac{\log(n)+\log(1+\frac 1n)}{\log(n)}\times\frac{n^2 -3n}{n^2-n-2}$$ So $\frac{u_{n+1}}{u_n}\to 1$ when $n\to \infty$ and the ratio test is inconclusive.

So, let us try Raabe's test $$n\left(\frac{u_n}{u_{n+1}}-1\right)=n\left(\frac{\log(n)}{\log(n+1)}\times\frac{n^2-n-2}{n^2 -3n}-1\right)$$ and use Taylor expansion for large $n$ (using the same trick as above for $\log(n+1)$). You should end with $$n\left(\frac{u_n}{u_{n+1}}-1\right)=\left(2-\frac{1}{\log \left({n}\right)}\right)+\frac{6-\frac{5}{2 \log \left({n}\right)}}{n}+O\left(\frac{1}{n^2}\right)$$ which tends to $2$ when $n\to \infty$. So, absolute convergence.

Answer 6

There is the upper bound: $$\sum_{n=4}^\infty\dfrac{\ln n}{n^2-3n} = \dfrac14\left(\sum_{n=4}^\infty\dfrac{\ln n}{n-3}-\sum_{n=4}^\infty\dfrac{\ln n}{n}\right) = \dfrac14\left(\ln4+\dfrac12\ln5+\dfrac13\ln6+\sum_{n=4}^\infty\dfrac{\ln(n+3)}{n}-\sum_{n=4}^\infty\dfrac{\ln n}{n}\right) = \dfrac14\left(\ln4+\dfrac12\ln5+\dfrac13\ln6+\sum_{n=4}^\infty\dfrac1n\ln\dfrac{n+3}{n}\right) = \dfrac14\left(\ln4+\dfrac12\ln5+\dfrac13\ln6+\sum_{n=4}^\infty\dfrac1n\ln\left(1+\dfrac3n\right)\right)\\< \dfrac14\left(\ln4+\dfrac12\ln5+\dfrac13\ln6+3\sum_{n=4}^\infty\dfrac1{n^2}\right).$$ Using well-known formula $$\sum_{n=1}^\infty\dfrac1{n^2}=\dfrac{\pi^2}6,$$ easy to get

$$\sum_{n=4}^\infty\dfrac{\ln n}{n^2-3n}<\dfrac14\left(\dfrac16\ln4^65^36^2+3\left(\dfrac{\pi^2}6-1-\dfrac14-\dfrac19\right)\right)\approx0.91.$$ So the sum converges, as increasing and bounded above function.