# Compare two below natural numbers: $2016^{2017} < 2017^{2016}$ [closed]

Help me Compare the two following natural numbers below $$2016^{2017} < 2017^{2016}?$$

Many thanks.

## closed as off-topic by choco_addicted, Thomas Shelby, stressed out, Kemono Chen, uniquesolutionFeb 22 at 19:15

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• Have you tried it yourself? Where are the arguments that should answer the question? – Parcly Taxel Sep 30 '16 at 14:33
• I tried but I can't. – Bích Hải Triều Sinh Sep 30 '16 at 14:34
• Please, explain what you tried and tell where you are stuck. It would be difficult to help you not knowing. Cheers and, by the way, welcome to this fantastic site. It is a very interesting problem. – Claude Leibovici Sep 30 '16 at 14:36
• I don't know which number is greater than? – Bích Hải Triều Sinh Sep 30 '16 at 14:39
• Let me give you a hint: You can try to prove an equivalent inequality $(1+1/2016)^{2016}<2016$, and try to estimate an upper bound for $(1+1/n)^n$ for $n\in\mathbb N_+$. – Yai0Phah Sep 30 '16 at 15:01

Taking logarithms of both sides we get: $$2016^{2017} > 2017^{2016}\Leftrightarrow 2017\ln 2016>2016\ln 2017 \Leftrightarrow \\ {} \\ \Leftrightarrow \frac{\ln 2016}{2016}>\frac{\ln 2017}{2017}$$ The last relation is true because the function $f(x)=\frac{\ln x}{x}$ is (why ?) strictly decreasing for $x>e$.
We make a fraction for the two numbers. So we have: $$\frac{2016^{2017}}{2017^{2016}}=\frac{2016.2016.2016...}{2017.2017.2017...}=2016(1-\frac{1}{2017})(1-\frac{1}{2017})...=\frac{2016}{e}>1$$. In a nutshell, we have $$2016^{2017}>2017^{2016}$$
Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.