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I'm having some more trouble with induction I know how to prove this using $\ln$, but I need to use induction only.

prove that:

$(n+1)^n < n^{n+1}$

for any $ n\ge 3$

Martin Sleziak
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Tal
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    Induction would only work for $n$ an integer...(you would need to follow the induction with an explanation of how it proves the result.) – user1729 Aug 04 '14 at 14:36
  • http://math.stackexchange.com/questions/600640/induction-nn1-n1n-and-n2-leq-left-fracn-12n-16 –  Aug 04 '14 at 14:43
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    This is in fact true for $n \geq e$, where $x/\log(x)$ is increasing on $(e,\infty)$. Why use induction here? – Christopher K Aug 04 '14 at 15:05
  • See also http://math.stackexchange.com/questions/77935/prove-by-induction-that-for-all-n-geq-3-nn1-n1n and http://math.stackexchange.com/questions/1011035/proving-that-n-1nnn-1 – Martin Sleziak Oct 18 '15 at 21:00

4 Answers4

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You need only to show: $$\frac{(n+1)^n}{n^{n+1}}>\frac{(n+2)^{n+1}}{(n+1)^{n+2}}.$$ But this follows from $$ (n+1)^{2n+2}=(n^2+2n+1)^{n+1}>(n(n+2))^{n+1}. $$

Chen Jiang
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Hypothesis \[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\] Basis \[ 3^{3+1} > (3+1)^{3} \] \[ 3^{4} > 4^{3} \] \[ 81 > 64\] Induction \[ (n+1)^{n+2}=n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] \] Now via the induction hypothesis, we have \[ n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] > (n+1)^{n}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] \] $$ \begin{align} (n+1)^{n}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] &= (n+2)^{n+1}\left[\frac{(n+1)^{2n+2}}{n^{n+1}(n+2)^{n+1}}\right]\\&= (n+2)^{n+1}\left[\frac{(n^{2}+2n+1)^{n+1}}{(n^{2}+2n)^{n+1}}\right] \end{align} $$ Since $n$ is positive and the numerator is larger than the denominator, then the fraction is greater than $1$. Hence, \[ (n+2)^{n+1}\left[\frac{(n^{2}+2n+1)^{n+1}}{(n^{2}+2n)^{n+1}}\right] > (n+2)^{n+1} \] Thus, \[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\] I hope this helps you understand.

robjohn
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k170
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You can also use the Bernoulli inequality $1+na\leq (1+a)^n$ which is valid for $-1\leq a$

write your inequality as

$$\frac{n+1}{n^2} < \frac{n^{n-1}}{(n+1)^{n-1}}=\left(1-\frac{1}{n+1}\right)^{n-1}$$

now by Bernoulli

$$\frac{2}{n+1} =1-\frac{n-1}{n+1}\leq \left(1-\frac{1}{n+1}\right)^{n-1}$$ so it remains to show $$\frac{n+1}{n^2} <\frac{2}{n+1}, $$ which is equivalent with $2< (n-1)^2$ and this is true for $n\geq 3$.

Rene Schipperus
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Taking natural logarithm, you will get $$ \frac{\ln(n+1)}{n+1}<\frac{\ln n}{n}. \tag{1}$$ Let $f(x)=\frac{\ln x}{x}$ for $x\ge 1$. Then $f'(x)=\frac{1-\ln x}{x^2}<0$ for $x>1$, namely $f(n)$ is decreasing for $n>1$. Thus $f(n+1)<f(n)$ which is (1).

xpaul
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