Show by induction: $(1+\frac{1}{n})^{n}<n$

Question

Show by induction that for all natural numbers n>3

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

Let $(1+\frac{1}{n})^n<n$ be true !

We show that $(1+\frac{1}{n+1})^{n+1})<n+1$

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

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

So now we just need to show that

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

So here , If I show that :

$(1+\frac{1}{n+1})<(1+\frac{1}{n})$

Does that mean im done with exercise ?

Answer 1

HINT: we have to prove that $\left(1+\frac{1}{n+1}\right)^{n+1}<n+1$ this is equivalent to $$\frac{\ln(n+2)}{n+2}<\frac{\ln(n+1)}{n+1}$$ this is true for $n>3$ for the proof consider the function $$f(x)=\frac{\ln(x)}{x}$$ with the extremum (maximum) $1/e$ at $x=e$

Answer 2

You have to prove for $n=4$ in order that mathematical induction work! To show the desired implication, note that if $n \geq 4$ is such that $$ \bigg( 1 + \frac{1}{n} \bigg)^{n} < n, $$ then $$ \bigg( 1 + \frac{1}{n+1} \bigg)^{n+1} = \bigg( 1 + \frac{1}{n+1} \bigg)^{n} \bigg( 1 + \frac{1}{n+1} \bigg) < \bigg( 1 + \frac{1}{n} \bigg)^{n}\bigg(1 + \frac{1}{n+1} \bigg) < n + \frac{n}{n+1} < n+1. $$

Note also that this implication makes no use of the hypothesis that $n \geq 4$, so if you do not check the initial proposition then a wrong conclusion could be drawn!