Prove that $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence Prove that the sequence $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence for $n>2.$
We have to show that $f(n+1)<f(n)$ for all $n>2.$ We have
$$
f(n+1)-f(n)=((n+1)!)^{\frac{1}{n+1}}-\frac{n+2}{2}-\left((n!)^{\frac{1}{n}}-\frac{n+1}{2}\right)=((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}-\frac{1}{2} <0,
$$
and reduce the problem to the following inequality
$$
((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}<\frac{1}{2}.
$$
With Maple I calculate 
$$
\lim\limits_{n \to \infty}(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}})=\frac{1}{e}<\frac{1}{2},
$$
but in is not enought to prove the inequality. Any ideas?
 A: We have 
$$\begin{align}\frac{\left(\frac12+n!^{1/n}\right)^{n+1}}{(n+1)!}&=\frac{(n!^{1/n})^{n+1}+(n+1)\cdot\frac12(n!^{1/n})^{n}+{n+1\choose2}\cdot\frac14(n!^{1/n})^{n-1}+\ldots}{(n+1)!}
\\&>\frac{n!^{1/n}}{n+1}+\frac12+\frac n{8\cdot n!^{1/n}}\end{align}$$
Using $a x+\frac bx\ge 2\sqrt{ab}$ (arithmetic-geometric inequality), we find that for $n\ge 2$
$$\frac{\left(\frac12+n!^{1/n}\right)^{n+1}}{(n+1)!}>\frac12+\sqrt{\frac{n}{2(n+1)}} \ge \frac12+\sqrt{\frac{1}{3}}>1$$
so that
$$ \frac12+n!^{1/n}>(n+1)!^{1/(n+1)}$$
and the claim follows.
A: Possible Proof by Induction.
Base case $n =3$, $$f(3) = (3!)^\frac{1}{3} -\frac{3 + 1}{2}$$
and so $$f(3) = (6)^\frac{1}{3} - 2  \approx -0.18$$ and now
$$f(4) = (4!)^\frac{1}{4} -\frac{4 + 1}{2}$$
and so $$f(4) = (24)^\frac{1}{4} -2.5 \approx -2.9$$
and so we have $f(4) < f(3)$ and so the base case works. Now assume it works for a value $k > 2$ that is $f(k+1) < f(k)$ now we try to prove that $f(k+2) < f(k+1)$ 
Given that $f(k+1) < f(k)$ we have that $$\large{(k!)^\frac{1}{k} -\frac{k+1}{2} > (k+1)!^\frac{1}{k+1} - \frac{k+2}{2}}$$
Now if we are able to deduce from here that $f(k+2) < f(k+1)$ then the answer is completed
Notice that 
$$f(k+2) = (k+2)!^\frac{1}{k+2} - \frac{k+3}{2}$$
