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I have derived the inequality if $k>1$, ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k<{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$

But, my problem is how to use this inequality to prove that if $n\geq1$, $(1+{1\over n})^n<(1+{1\over n+1})^{n+1}$

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use $AM-GM$,$$(1+\dfrac{1}{n})(1+\dfrac{1}{n})\cdots(1+\dfrac{1}{n})\cdot 1\le\left(\dfrac{n+1+1}{n+1}\right)^{n+1}$$

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  • $\begingroup$ i didn't expect it to so simple (+1) $\endgroup$ – Santosh Linkha Sep 29 '13 at 16:28
  • $\begingroup$ Yes, that's quite deviously clever. :-) $\endgroup$ – user43208 Sep 29 '13 at 16:35
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$$\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} = 1-\frac{1}{(n+1)^2} $$

By Bernoulli's inequality:

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

Putting it together:

$$\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n} \geq \left(1-\frac{n}{(n+1)^2}\right)\left(1+\frac{1}{n+1}\right) = 1+\frac{1}{(n+1)^3} $$

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