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March 6th, 2018, 11:10 AM   #1
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Show inequality true

Show that $\displaystyle (1+ \frac{1}{n} )^n < (1+ \frac{1}{n+1})^{n+1} \;$ for $\displaystyle n$ natural
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March 6th, 2018, 12:48 PM   #2
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Make an attempt and post your work. Are you expected to use any particular method, such as induction?
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March 8th, 2018, 12:21 PM   #3
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Since $\displaystyle (1+\frac{1}{n})^n=f(n)<f(n+1)$ we can use derivatives for $\displaystyle Y=(1+\frac{1}{x})^x $ or
define $\displaystyle f(n)$ increasing for $\displaystyle n\in N$
From $\displaystyle \frac{f(n+1)}{f(n)}>1$ use Bernoulli's inequality $\displaystyle (1+r)^x>1+xr$ with conditions
$\displaystyle [1+(-(n+1)^{-2}))]^{1+n}>1+(-(n+1)^{-2}\Rightarrow (1+\frac{1}{n})^n < (1+\frac{1}{n+1})^{n+1}$
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