Which is larger, $\sqrt[2015]{2015!}$ or $\sqrt[2016]{2016!}$?

Solution 1:

Starting with:

$$\sqrt[2015]{2015!}\mid\sqrt[2016]{2016!}$$

Raise each side to the power of $2015\cdot2016$:

$$2015!^{2016}\mid2016!^{2015}$$

Divide each side by $2015!^{2015}$:

$$2015!^{1}\mid2016^{2015}$$

Write it explicitly as:

$$\underbrace{1\cdot2\cdot\ldots\cdot2015}_{2015\text{ terms}}\mid\underbrace{2016\cdot2016\cdot\ldots\cdot2016}_{2015\text{ terms}}$$

Obviously, the RHS is larger.

Solution 2:

Look at the function: $$y(x)=\sqrt[x]{x!}=\left(\Gamma(x+1)\right)^{1/x}$$ It's a monotonically strictly increasing function, as: $$y'(x) = \frac{1}{x}\left(\Gamma(x+1)\right)^{1/x-1}\Gamma'(x+1)>0,\forall x>1$$ $\blacksquare$

As a side note: $$\lim_{x\to\infty}y(x)=\frac{1}{e}x$$ So you can expect the RHS to be lager by a factor of around $1+1/2015$.

Solution 3:

$$\log \left(2016!^{1/2016}\right) -\log \left(2015!^{1/2015}\right)=$$ $$=\frac {1}{2016}\log (2016)-\left(\frac {1}{2015}-\frac {1}{2016}\right)\sum_{n=1}^{2015}\log n=$$ $$=\frac {1}{2016}\log (2016)- \frac {1}{(2016)(2015)}\sum_{n=1}^{2015}\log n>$$ $$>\frac {1}{2016}\log (2016)-\frac {1}{(2016)(2015)}\sum_{n=1}^{2015}\log (2015)=$$ $$=\frac {1}{2016}\log (2016)-\frac {1}{2016}\log (2015)\;>0\;.$$