-1
$\begingroup$

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

Many thanks.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – choco_addicted, Thomas Shelby, stressed out, Kemono Chen, uniquesolution
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Have you tried it yourself? Where are the arguments that should answer the question? $\endgroup$ – Parcly Taxel Sep 30 '16 at 14:33
  • $\begingroup$ I tried but I can't. $\endgroup$ – Bích Hải Triều Sinh Sep 30 '16 at 14:34
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici Sep 30 '16 at 14:36
  • $\begingroup$ I don't know which number is greater than? $\endgroup$ – Bích Hải Triều Sinh Sep 30 '16 at 14:39
  • 2
    $\begingroup$ 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 $n\in\mathbb N_+$. $\endgroup$ – Yai0Phah Sep 30 '16 at 15:01
5
$\begingroup$

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 $x>e$.

$\endgroup$
  • $\begingroup$ @KonKan. So I think 2016^2017 > 2017^2016. Thanks. $\endgroup$ – Bích Hải Triều Sinh Sep 30 '16 at 14:42
  • $\begingroup$ @Bích Hải Triều Sinh, you are welcome. $\endgroup$ – KonKan Sep 30 '16 at 14:53
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.