Does $\sum{\ln n\over n^{4/3}}$ converge or diverge?

Question

Does $\displaystyle\sum{\ln n\over n^{4/3}}$ converge or diverge? Which test should I use? I tried the ratio test and root test but both of them are inconclusive. I could try the comparison test but I don't know which function I have to compare with.

Answer 1

By the integral test you may compare your series with the following integral: $$ \int_1^\infty \frac{\ln x}{x^{4/3}} dx=\left.\frac{x^{-4/3+1}}{-4/3+1}\ln x\right|_1^\infty-\frac1{-4/3+1}\int_1^\infty x^{-4/3+1}\frac1{x} dx=9<\infty $$ giving the convergence of your series $$ \sum_1^\infty\frac{\ln n}{n^{4/3}}. $$

Answer 2

This may require a little computation. Notice that if you can compare $\frac{\ln(n)}{n^{4/3}}$ to a converging $p$-series, then the answer will be obvious. Thus proceed as follow: $$\frac{\ln(n)}{n^{4/3}} = \frac{1}{n^{7/6}}\frac{\ln(n)}{n^{1/6}} <\frac{1}{n^{7/6}}$$ This is only true if $n$ is larger than about $2.5\times 10^7$. But you can divide the sum into two: $$\sum_{n=1}^{n = 2.5\times 10^7}\frac{\ln(n)}{n^{4/3}} + \sum_{n=2.5\times 10^7}^{\infty}\frac{\ln(n)}{n^{4/3}}$$ The first term converges since it is a finite sum, while the second converges because it is less than the $p$-series $\frac{1}{n^{7/6}}$. Not very useful on exam unless you realize that for there is always a large value of $n$ that can force the inequality in the beginning.

Answer 3

An other way is to compare $\ln(.)$ with $\sqrt{.}$ in fact $$\forall x \in \mathbb{R}, \ln(x) \le \sqrt{x}$$

Then $$\forall n \in \mathbb{N}, \ln(n) \le \sqrt{n}$$

So for $n> 1$ $$\left|\frac{\ln(n)}{n^{\frac{4}{3}}}\right|\le\left|\frac{\sqrt{n}}{n^{\frac{4}{3}}}\right|=\frac{1}{n^{\frac{6}{5}}}$$

And obviously $\frac{6}{5}>1$ so the series is convergent.

Answer 4

Let's apply Cauchy condensation test :

$$2^n a_{2^n}=2^n\cdot \frac{\log 2^n}{2^{4n/3}}=\log{2}\cdot \frac{n}{2^{n/3}}$$

$\sum_n^\infty \frac{n}{2^{n/3}} $ clearly converges, so the given series also converges.