5
$\begingroup$

$\displaystyle(1 + \frac{1}{n})^n < n$ for $n \gneq 3$

yes for $n = 1$ it is true

I assume it is true for $n = k$ and get

$\displaystyle(1+\frac{1}{k})^k < k$

I then go to $\displaystyle(1 +\frac{1}{k+1})^{k+1} < k+1$ and now I spend an hour doodling.

  • 0
    I've latex-ed up your post. I wasn't entirely sure how to interpret n>/=3. I think it should be $n \gneq 3$, but I wasn't entirely sure. Is this correct? (It makes sense for the problem, etc.)2011-12-08
  • 0
    Thanks for latexing it up. This is my first time on this site!2011-12-08
  • 5
    Actually, much more is true: $\left(1 + \dfrac{1}{n}\right)^n < 3$.2011-12-08
  • 0
    @Tyson: just put dollar signs around your maths, and use curly brackets when you want latex to group things together. For example, a^{k+1} looks like $a^{k+1}$ while a^(k+1) looks like $a^(k+1)$.2011-12-08
  • 0
    Thanks let me try $a^{k+1}$2011-12-08
  • 4
    (Abstract) duplicate of [Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$](http://math.stackexchange.com/questions/77935/prove-by-induction-that-for-all-n-geq-3-nn1-n1n) (just divide both sides by $n^n$)2011-12-08
  • 0
    Now let n go to infinity.2011-12-08
  • 0
    $\left(1+\frac1n\right)^n < e = 2.7182...$ for $n>0$2011-12-08

2 Answers 2