What is the proper convergence proof for this sequence? Limit of $n^{1/2}(n^{1/n}-1)=0$, as $n$ approaches infinity.
I need a strict math proof that this sequence converges to zero, but without using L'Hôpital's rule, because I am not allowed to use it yet. Thanks in advance.

Denote $a_n=n^\tfrac{1}{n}-1, \;\;  b_n={\sqrt{n}\left(n^\tfrac{1}{n}-1\right)}.$
Obviously, $a_n \geqslant {0}.$  Rewrite $n^{\tfrac{1}{n}}=1+a_n $ and apply the binomial formula
\begin{gather}n=(1+a_n)^n= \\
=1+na_n+\dfrac{n(n-1)}{2!}a_n^2+\dfrac{n(n-1)(n-2)}{3!}a_n^3+\ldots+a_n^n \geqslant \\
\geqslant \dfrac{n(n-1)(n-2)}{3!}a_n^3.
\end{gather}
For $n\geqslant{4}$
$$n-1 \geqslant \dfrac{n}{2}, \\
n-2 \geqslant \dfrac{n}{2}.
$$
Therefore, $$n \geqslant \dfrac{n^3}{4 \cdot 3!}a_n^3=\dfrac{n^3a_n^3}{24} {\;\;}\Rightarrow  {\;\;}a_n^3 \leqslant \dfrac{24}{n^2}.$$
Then ${a_n}\leqslant \dfrac{{\sqrt[3]{24}}}{n^{\tfrac{2}{3}}}$ and $${0} \leqslant{b_n} =\sqrt{n}a_n \leqslant \sqrt{n}\dfrac{\sqrt[3]{24}}{n^{\tfrac{2}{3}}}={\sqrt[3]{24}}n^{-\tfrac{1}{6}} \underset{n\to\infty} {\to}  {0}.$$
