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I know this seems trivial, but how could I proof this? Should I use Induction? Where $n$ is an integer.

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  • $\begingroup$ $x^x=e^{x\ln x}$ is increasing on $[1,\infty[$ $\endgroup$ – Hippalectryon Nov 12 '15 at 21:07
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    $\begingroup$ We have $n\le n+1$. So $n^k\le (n+1)^k$ for every positive integer $k$. $\endgroup$ – André Nicolas Nov 12 '15 at 21:11
  • $\begingroup$ use binomial expansion. $\endgroup$ – Boltzee Nov 12 '15 at 21:11
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    $\begingroup$ Proving such claims is like shooting rockets on ants. For $n>0$, the equal-sign can be omitted. But if you do want a proof, you can do it with $n<n+1\ \rightarrow \ ln(n)<ln(n+1)\ \rightarrow \ n ln(n) < n ln(n+1)\ \rightarrow\ n^n<(n+1)^n$ for $n>0$. $\endgroup$ – Peter Nov 12 '15 at 21:17
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    $\begingroup$ $0^0$ is usually defined to be $1$, so $0^0=1^0=1$. $\endgroup$ – Peter Nov 12 '15 at 21:22
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Using the binomial theorem,

$$ (n+1)^n=\sum_{i=0}^n\binom{n}{i}n^i. $$

One of the terms is $n^n$ and the rest are positive, so $(n+1)^n$ is $n^n+$something positive, and the inequality follows.

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    $\begingroup$ to prove a simple statement by use a rather complex one seems not a very useful idea to me. $\endgroup$ – miracle173 Nov 12 '15 at 21:49
  • $\begingroup$ @miracle173 Agreed, but this method doesn't require monotonicity of the power function to prove. $\endgroup$ – Michael Burr Nov 13 '15 at 16:59
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$$ (n+1)^n = [n(1+1/n)]^n = n^n(1+1/n)^n \geq n^n\cdot1^n = n^n $$

whenever $n \in \mathbb{N}$ is greater than or equal to $1$.

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  • $\begingroup$ Dang. Beat me to it by mere seconds. Good job. $\endgroup$ – Rick Decker Nov 12 '15 at 21:12
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    $\begingroup$ and why $(1+1/n)^n \geq 1^n$? Because exponentiation is an increasing function? Then $(n+1)^n \geq n^n$ follows immediately. $\endgroup$ – miracle173 Nov 12 '15 at 21:18
  • $\begingroup$ @miracle173: That's an admissible point. I would say that provided $n$ is a natural number, you can then show this secondary fact by induction: $1+1/n \geq 1$, and then $(1+1/n)^k = (1+1/n)(1+1/n)^{k-1} \geq (1+1/n)^{k-1}$. But if $n$ is in general a real, then I would agree with you that the above is not a sufficient demonstration. $\endgroup$ – Brian Tung Nov 12 '15 at 22:09
  • $\begingroup$ to prove $(1+1/n)^n \geq 1^n$ by induction is almost the same as to prove $n^{n} \leq {(n+1)}^{n}$. So this is not a proof at all. But it seems that some readers are excited by the formulas $\endgroup$ – miracle173 Nov 12 '15 at 23:40
  • $\begingroup$ @miracle173: Yeah, I guess. I'm assuming that this appeals to the intuition of some people that the more direct approach does not. (I.e., it's easier to see that $(1+1/n)^n > 1$ than the other). I concede I'm occasionally perplexed by what gets votes here. $\endgroup$ – Brian Tung Nov 12 '15 at 23:43
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From Andre's hint in the comments:

$$n \leq n + 1$$

In fact, strict inequality holds:

$$n < n + 1$$

Assuming $n \in \mathbb{N}$ (since I see that you've mentioned induction), we can raise both sides of the last inequality to the $n$th power:

$$n^n < (n + 1)^n.$$

This implies that

$$n^n \leq (n + 1)^n,$$

since $\leq$ is to be interpreted as less than OR equal to.

QED

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Induction is a good and simple way to prove $$a \lt b \implies a^n<b^n$$ for $ a,b,n \in \mathbb{N}$ because $$a^{n+1}=a^n\cdot a<b^n \cdot b=b^{n+1}$$ From this your statement follows.

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