Existence of limits of these two sequence.

Question

Examine that these two sequence are divergent $$ \lim_{n\to\infty} \sqrt{n}(\sqrt[n]{n}-1) $$

$$ \lim_{n\to\infty} \sqrt[n]{3^{n}+(-3)^{n}+n^{-1}} $$

Answer 1

$$\sqrt{n}(\sqrt[n]{n}-1)=\sqrt{n}\frac{\ln n}{n}{e^{\frac{\ln n}{n}}-1\over \frac{\ln n}{n}}=\frac{\ln n}{\sqrt n}{e^{\frac{\ln n}{n}}-1\over \frac{\ln n}{n}}=\frac{2\ln \sqrt n}{\sqrt n}{e^{\frac{\ln n}{n}}-1\over \frac{\ln n}{n}}$$ but $$\lim_{n\to\infty}\frac{2\ln \sqrt n}{\sqrt n}=0\text{ & }\lim_{n\to\infty}{e^{\frac{\ln n}{n}}-1\over \frac{\ln n}{n}}=(e^x)'(0)=1$$ (Note that by cahngaing variable $h=\frac{\ln n}{n}$ we have $\lim_{n\to\infty}{e^{\frac{\ln n}{n}}-1\over \frac{\ln n}{n}}=\lim_{h\to o}\frac{e^h-e^0}{h}=(e^x)'(0))$. For

$$a_n=\sqrt[n]{3^{n}+(-3)^{n}+n^{-1}}=3\sqrt[n]{1+(-1)^{n}+\frac{1}{n3^n}}$$ We have $a_{2n+1}=\sqrt[n]{\frac{1}{n}}=\frac{1}{\sqrt[n]{n}}\Rightarrow\color{red}{\lim a_{2n+1}=1.}$ And $$ a_{2n}=3\sqrt[n]{2+\frac{1}{n3^n}}$$ then $$3\sqrt[n]{2}\leq a_{2n}\leq 3\sqrt[n]{3}$$ since $\lim_{n\to\infty}a^\frac{1}{n}, a>0$ thus $\color{red}{\lim a_{2n}=3.}$ Thus $a_n$ does not convergent.