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March 6th, 2018, 11:10 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27  Show inequality true
Show that $\displaystyle (1+ \frac{1}{n} )^n < (1+ \frac{1}{n+1})^{n+1} \;$ for $\displaystyle n$ natural

March 6th, 2018, 12:48 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond 
Make an attempt and post your work. Are you expected to use any particular method, such as induction?

March 8th, 2018, 12:21 PM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 
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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