The root test for the series

Question

I am reading an example of root test for a serie: $$\sum_{n=1}^\infty\frac{3n}{2^n}.$$ So applying the root test, we get $$\lim_{n \to \infty}\sqrt[n]{\frac{3n}{2^n}}=\lim_{n \to \infty}\frac{\sqrt[n]{3n}}{2}=\frac{1}{2}\lim_{n\to\infty}\exp(\frac{1}{n}\ln (3n))\\=\frac{1}{2}\exp(\lim_{n\to\infty}\frac{\ln (3n)}{n})=\frac{1}{2}\exp(\lim_{n\to\infty}\frac{\frac{3}{3n}}{1})=\frac{1}{2}e^0=\frac{1}{2}.$$

I don't understand the expression starting to where the $\exp$ is used. can someone explain which formula is used there? Then why $$\lim_{n\to\infty}\frac{\ln (3n)}{n}=\lim_{n\to\infty}\frac{\frac{3}{3n}}{1}?$$

Answer 1

This is just a way of rewriting things using the exponential. Notice that $$ \exp (\frac1n \log(3n))=\exp(\log(\sqrt[n]{3n}))=\sqrt[n]{3n}. $$ Here, $\log$ stands for your $\text{ln}$.

The final question is justified by L'Hôpital's rule, that is, since you have an indetermination of the type $\infty/\infty$, differentiating both numerator and denominator yields the computable limit. Notice that the derivative of $\log(3n)$ is $3/(3n)$ and the derivative of $n$ is $1$.

Answer 2

The definition of a (possibly irratioanl power of a positive number is $$x^y\stackrel{\text{def}}{=}\exp(y\ln x).$$ For you other question, it is a high-school result that $\lim_{x\to\infty}\dfrac{\ln x}x=0$ or, in terms of asymptotic analysis, $\;\ln x=_{x\to \infty} o(x) $.

More generally, for any $\alpha,\beta >0$, we have $$\lim_{x\to\infty}\frac{(\ln x)^\alpha}{x^\beta}=0.$$