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

This was a question in a maths contest, where no calculator was allowed. Also, note that only a (>,< or =) relationship is being searched for and not the value of the numbers itself.

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

What I've done: My approach is to divide one number by the other and infer from the result which number is the bigger one;
WolframAlpha gives $\frac{\sqrt[2016]{2016!}}{\sqrt[2015]{2015!}}=1.0049\ldots$, so clearly $\sqrt[2016]{2016!}>\sqrt[2015]{2015!}$

Let $a=\sqrt[2016]{2016!}$ and $b=\sqrt[2015]{2015!}$

$\therefore a=\sqrt[2016]{2016!}={2016!}^{1 \over 2016}=2016^{1 \over 2016}\times2015!^{1\over 2016}=\sqrt[2016]{2016}\cdot \sqrt[2016]{2015!}$
$\therefore b=\sqrt[2015]{2015!}={2015!}^{1 \over 2015}={2015!}^{\frac{2016}{2015}\cdot\frac{1}{2016}}=\sqrt[2016]{2015!^{2016 \over 2015}}$

Hence \begin{align} \require{cancel} \frac{a}{b}=\frac{\sqrt[2016]{2016!}}{\sqrt[2015]{2015!}}&=\frac{\sqrt[2016]{2016}\cdot \sqrt[2016]{2015!}}{\sqrt[2016]{2015!^{2016 \over 2015}}}\\ &=\sqrt[2016]{2016}\cdot \sqrt[2016]{2015!^{\frac{-1}{2015}}}= \cancelto{*}{\sqrt[2016]{\frac{2016}{2015!^{2015}}} \quad \text{which appears to be} <1}\\ =\sqrt[2016]{\frac{2016}{2015!^\frac{1}{2015}}}\\ \end{align} That is $\cancelto{*}{\frac{a}{b}<1 \implies a<b}$ which is false as per the result from WA.

EDIT:
*: Correction due to error pinpointed out by Daniel Fischer. But now I'm stuck; how do I infer what value is $\sqrt[2016]{\frac{2016}{2015!^\frac{1}{2015}}}$?

So, where did I go wrong?. How do I proceed now?

• $$(2015!)^{\frac{-1}{2015}} = \frac{1}{(2015!)^{\frac{1}{2015}}} \neq \frac{1}{(2015!)^{2015}}$$ Jan 4 '16 at 10:17
• It's easier to work with only integer exponents. You have $$\sqrt[n+1]{(n+1)!} > \sqrt[n]{n!} \iff ((n+1)!)^n > (n!)^{n+1}.$$ I think you can work from that. Jan 4 '16 at 10:40
• But how do we know that $\sqrt[n+1]{(n+1)!} > \sqrt[n]{n!}$ ? Jan 4 '16 at 10:43
• $$\frac ab=\sqrt[2015\cdot 2016]{\frac{2016^{2015}}{2015!}}\gt 1$$ Jan 4 '16 at 10:46

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.

• | looks like dividing relation so it is little bit confusing. Jan 4 '16 at 11:57
• "Raise each side to the power of $2015⋅2016$" nice trick Jan 4 '16 at 12:17
• @barakmanos Well, how about $\gtreqless$ or $\stackrel{?}{\gtreqless}$ ? Jan 4 '16 at 12:28
• @dust05: I wouldn't even know how to make those in LaTex. Perhaps I should just add "vs" in between the expressions... Or if I could make that "$\mid$" longer... Jan 4 '16 at 13:12
• @barakmanos Simply begin with the assumption that $\sqrt[2015] {2015} > \sqrt[2016] {2016}$ and then come to the conclusion that $\sqrt[2015] {2015} \not > \sqrt[2016] {2016}$. No fancy notation needed. Jan 5 '16 at 5:34

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 $1+1/2015$.

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