How to prove (without differentiation) that $\exists n_0 \in \mathbb{N}$ such that $n^{\frac{1}{2n}}$ is strictly monotonically decreasing? 
Let $f(n) = n^{\frac{1}{2n}}$. 
Prove the existence of an $n_0 \in \mathbb{N}$ such that $f(n) > f(n+1) \forall n \geq n_0$ and find its smallest possible value.

None of my attempts to solve this problem, namely a proof by induction and a proof by contradiction, have been fruitful. However, I did find that $\lim_{n \to \infty} (f(n)) = 1$, since $\lim_{n \to \infty} \left(\frac{1}{2n}\right) = 0$.
What's an appropriate approach to prove the statement, ideally without the use of derivatives?
 A: A quite elementary way is as follows:


*

*Note that $x^{\frac{1}{2}}=\sqrt{x}$ is strictly increasing.

*$\Rightarrow$ It is enough to show $n^\frac{1}{n}> (n+1)^\frac{1}{n+1}$ from a certain $n_0 \in \mathbb{N}$ on


You have
$$n^\frac{1}{n}> (n+1)^\frac{1}{n+1} \Leftrightarrow n^{n+1} > (n+1)^n$$
$$\Leftrightarrow n > \left(1+\frac{1}{n} \right)^n$$
You either know that the expression on the RHS approaches $e$ from below or you show it for $n\geq n_0=3$ using the binomial expansion:
$$\left(1+\frac{1}{n} \right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\frac{1}{n^k}$$ $$\leq 2+ \sum_{k=2}^n\frac{1}{2^{k-1}}< 2+\sum_{k=1}^{\infty}\frac{1}{2^{k}}=3$$
A: That is  because the function $f(x)=x^{\tfrac 1{2x}}$ is ultimately decreasing. To prove it, observe it is equivalent to proving it for the function $\;g(x)=\ln(f(x))=\dfrac1{2x}\ln x$. Now 
$$g'(x)=\frac 12\biggl(-\frac1{x^2}\ln x+\frac1{x^2}\biggr)=\frac{1-\ln x}{2x^2}.$$
Do you see it now?
A: 
In this development, we forgo the use of calculus and rely on elementary analysis only.  We now proceed.


Let $a_n=n^{1/2n}$.  To show that $a_n$ is decreasing, it suffices to show that $\frac{a_{n+1}}{a_n}<1$.  This is equivalent to our showing that $2\log\left(\frac{a_{n+1}}{a_n}\right)=\log\left(\frac{a_{n+1}}{a_n}\right)^2<0$.  
Using $\log(x)\le x-1$, we see that 
$$\begin{align}
\log\left(\frac{a_{n+1}}{a_n}\right)^2&=\frac1{n+1}\log(n+1)-\frac1{n}\log(n)\\\\
&=\frac1{n(n+1)}\left(n\log(n+1)-(n+1)\log(n) \right)\\\\
&=\frac1{n(n+1)}\left(n\log(n)+n\log\left(1+\frac1n\right)-(n+1)\log(n) \right)\\\\
&=\frac1{n(n+1)}\left(n\log\left(1+\frac1n\right)-\log(n) \right)\\\\
&\le \frac1{n(n+1)}\left(1-\log(n) \right)\\\\
&<0
\end{align}$$
whenever $\log(n)>1$.  This occurs whenever $n>3>e$.  And we are done!
A: The function $f(x)=x^{\frac{1}{2x}}=e^{\frac{\log{x}}{2x}}$ is  strictly decreasing if $x>e$.
You can prove this by differentiation.
So  $n=  [e]+1$ is the minimun.
Here $[x]$ denotes the integer part of $x$
