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.
 A: 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 $${0} \leqslant{b_n} =\sqrt{n}a_n \leqslant \sqrt{n}\dfrac{\sqrt[3]{24}}{n^{\tfrac{2}{3}}}={\sqrt[3]{24}}n^{-\tfrac{1}{6}} \underset{n\to\infty} {\to}  {0}.$$
A: Note: This recapitulates work I did a few years ago.
From Bernoulli's inequality, if $k > 1$,
$(1+n^{-1+1/k})^n \ge 1+ n^{1/k} > n^{1/k}$.
Raising to the $k/n$ power,
$(1+n^{-1+1/k})^k > n^{1/n}$.
To bound the left side of this inequality,
we need what I call a contra-Bernoulli inequality
(CBI),
which provides an upper bound to $(1+x)^k$.
Any such bound works only for a bounded range of $x$,
for if $(1+x)^k < 1+c x$,
then $1+cx > (1+x)^k > 1+x^k$,
so $x < c^{1/(k-1)}$.
I have come up with two of these CBIs (and will leave the proofs to the reader):


*

*If $0 < x < 1/k$ then $(1+x)^k < 1/(1-kx)$.

*if $0 < x < M$, then $(1+x)^k < 1+c(k,M)x$, where $c(k,M) = ((1+M)^k-1)/M$.
CBI 1 gives us  (after some work) 
$n^{1/n} < 1+2kn^{-1+1/k}$ if $n > (2k)^{k/(k-1)}$.
CBI 2 gives us
this: if $n^{-1+1/k} < M$, then $n^{1/n} < 1+((M+1)^k-1)/M)n^{-1+1/k}$.
Letting $M = 1$, this becomes
$n^{1/n} < 1+(2^k-1)n^{-1+1/k}$.
For the OP's query, choose any $k > 2$.
If $k = 3$,
from the CBI 2 result,
$n^{1/n} < 1+7n^{-2/3}$,
so $n^{1/n} - 1<7n^{-2/3}$
or $n^{1/2}(n^{1/n} - 1)<7n^{-1/6}$
which obviously goes to zero.
What I like about these results is that
they are completely elementary - no calculus, limits,
logarithms, or other analysis is used.
All that is needed is
Bernoulli's inequality
(which is easily proved by induction)
and the existence of the $n^{th}$ root
of any positive real for any positive integer $n$.
Note that this shows that
$n^{1/n} < 1+c(k)n^{-1+1/k}$,
where $c(k)$ is explicitly computable,
and can be bounded nicely
by taking $n$ large enough
compared with $k$.
If you choose $1/k < \epsilon$,
then
$n^{1-\epsilon}(n^{1/n}-1)
< c(k)n^{1/k-\epsilon}
$,
so $n^{1-\epsilon}(n^{1/n}-1) \to 0$
for any $\epsilon > 0$.
Not too bad for a completely elementary argument.
I have submitted this over a year ago the a MAA publication,
but have not heard anything.
I am glad to take advantage of this opportunity.
A: The convexity of the exponential function implies that
$$
e^t-1\leq2t,\ \ t\in[0,1]
$$
For $n$ big enough we have $\frac1n\log n\leq1$, since $\frac1n\log n\to0$ as $n\to\infty$. Then
$$
n^{1/2}(n^{1/n}-1)=n^{1/2}\,(e^{\frac1n\,\log n}-1)\leq n^{1/2}\,\frac2n\log n=\frac2{n^{1/2}}\,\log n\to0.
$$
