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I have to show that

$a_n=\left(1+\frac{1}{n}\right)^n$

is monotonic increasing. I struggle alot and tryng to find a right form to epxress that,

$a_n\le a_{n+1}$

a little hint what might help would be great. Im in my first year in the introduction to real analysis, so tipps with derivative or more advanced arent usefull :/

I really only like to get a little hint :)

CoffeeArabica
  • 991
  • $$a_n \le a_{n+1} \iff (n+1)^{2n+1} \le n^n(n+2)^{n+1} \iff \frac{n+1}{n+2} \le \left( 1-\frac{1}{(n+1)^2} \right)^n$$ Besides, $$\left( 1-\frac{1}{(n+1)^2} \right)^n \ge 1-\frac{1}{(n+1)^2}n $$ and clearly $$1-\frac{1}{(n+1)^2}n \ge 1-\frac{1}{n+2}$$ Done. – Paresseux Nguyen Dec 04 '20 at 00:12
  • Try and show that $\frac{a_{n+1}}{a_n}\geq1$ – Matthew H. Dec 04 '20 at 00:18

2 Answers2

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Let $$ f(x)=(1+\frac1x)^x $$ and show $f(x)$ is increasing.

xpaul
  • 44,000
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$$\frac{a_{n+1}}{a_n}=\frac{\left( 1+ \frac{1}{n+1} \right)^{n+1}}{\left( 1+ \frac{1}{n} \right)^{n}}=\frac{n+1}{n}\left( 1- \frac{1}{(n+1)^2} \right)^{n+1}>\\ >\frac{n+1}{n}\left( 1- \frac{1}{(n+1)} \right)=1$$

zkutch
  • 13,410