Help me Compare the two following natural numbers below $$2016^{2017} < 2017^{2016}?$$
Many thanks.
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Martin Sleziak
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Bích Hải Triều Sinh
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3Have you tried it yourself? Where are the arguments that should answer the question? – Parcly Taxel Sep 30 '16 at 14:33
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I tried but I can't. – Bích Hải Triều Sinh Sep 30 '16 at 14:34
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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
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I don't know which number is greater than? – Bích Hải Triều Sinh Sep 30 '16 at 14:39
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2Let 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
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@ClaudeLeibovici Our criterion is sometimes harsher when the problem seems more elementary. Those posts about moderner mathematics (such as functional analysis or algebraic geometry) without explaining attempts aren't that downvote- or closing-attractive. – Yai0Phah Sep 30 '16 at 15:06
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You are asking about inequality of the form $n^{n+1}<(n+1)^n$, which is equivalent to $\sqrt[n]n<\sqrt[n+1]{n+1}$. You can find several posts about this on this site. I have mentioned some of them in this answer. Or you can try searching in approach0 for the first inequality or the second inequality or for another equivalent form. – Martin Sleziak Dec 16 '16 at 10:14
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Do you want how to do the comparison or just the answer? – tc216 Dec 16 '16 at 23:45
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Also related: Fastest way to check if $x^y > y^x$? (and other posts linked there.) – Martin Sleziak Feb 22 '19 at 17:05
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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 $x>e$.
KonKan
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I don't know whether it's right or wrong but I still try as hard as possible.
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 $$2016^{2017}>2017^{2016}$$
Moreover, we easily see that if 0 < x <= 2, then the numerator is less than the denominator.
Bích Hải Triều Sinh
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Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.
lhf
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