Does $\sum{\ln n\over n^{4/3}}$ converge or diverge? 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.
 A: 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
$$
\sum_1^\infty\frac{\ln n}{n^{4/3}}.
$$
A: 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.
A: 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.
A: $$n\ge4\implies\ln n<n^{1/6}$$ so that the series is bounded by a convergent Zeta series.
A: 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.
