About the sequence $n^{\frac{1}{n}} -1$ I am having trouble with this multiple-select mcq question:

The value of $n^{\frac{1}{n}} -1$ 
  
  
*
  
*tends to $0$ as $n\to \infty$,
  
*is greater than $\frac{\log n}{n}$ for all $n\geq 3$,
  
*is greater than $\log n$ for all $n\geq 3$,
  
*is greater than $\frac{1}{\sqrt{n}}$ for all $n\geq 3$. 
  

I know that option 1 is correct, because $\lim_{n\to \infty}n^{\frac{1}{n}} =1$.
Also,
$$n^{\frac{1}{n}} -1=-1+ e^{\frac{\log n}{n}} =\frac{\log n}{n} + \frac{(\log n)^2}{n^2}/2! +...\geq \frac{\log n}{n},$$ meaning option 2 is also correct.
But I don't see how to prove or disprove the other two options.
Any help would be appreciated!
 A: To tackle the fourth question, you can let
$$f(x)=x^\frac{1}{x} - \frac{1}{\sqrt{x}}.$$
For $x^\frac{1}{x}$, it is easy to show that it is increasing by looking at its derivative, so the function $f(x)=x^\frac{1}{x} - \frac{1}{\sqrt{x}}$ is increasing. Then what you need to do is to check the case where $n=3$.
A: *

*Using the binomial theorem,
$$
\begin{align}
\left(1+\sqrt{\frac2n}\right)^n
&\ge1\color{#AAA}{+n\sqrt{\frac2n}}+\frac{n(n-1)}2\frac2n\color{#AAA}{+\cdots}\\
&\ge n\color{#AAA}{+\sqrt{2n}+\cdots}
\end{align}
$$
Therefore, $1\le n^{1/n}\le1+\sqrt{\frac2n}$. thus,
$$
\lim_{n\to\infty}\left(n^{1/n}-1\right)=0
$$


*Using the Taylor series for $e^x$,
$$
\begin{align}
n^{1/n}
&=e^{\log(n)/n}\\
&=1+\frac{\log(n)}n+\frac12\left(\frac{\log(n)}n\right)^2+\frac16\left(\frac{\log(n)}n\right)^3+\cdots
\end{align}
$$
Therefore,
$$
n^{1/n}-1\ge\frac{\log(n)}n
$$


*Since 1. is true, 3 cannot be.


*Bernoulli's Inequality says $\left(1+\frac1{\sqrt{n}}\right)^{\sqrt{n}}\ge2$. Because $\frac{\log(x)}x$ is decreasing for $x\gt e$, for $n\ge16$, $\frac{\log\left(\sqrt{n}\right)}{\sqrt{n}}\le\frac{\log(2)}2$; that is, $n\le2^{\sqrt{n}}$. Thus,
$$
\begin{align}
n^{1/n}
&\le2^{1/\sqrt{n}}\\
&\le1+\frac1{\sqrt{n}}
\end{align}
$$
Therefore, for $n\ge16$,
$$
n^{1/n}-1\le\frac1{\sqrt{n}}
$$
