# Another proof by induction

1. Oct 16, 2004

### quasar987

Basically I will have won if I can show that

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

can I? Not that I haven't tried...

This is also equivalent to showing that

$$n^n \geq (n+1)^{n-1}$$

Last edited: Oct 16, 2004
2. Oct 16, 2004

### quasar987

I found it. It wasn't so easy but it turns out that the sequence on the right has 3 for an upper bound. More precisely, it has e, the neperian number, has its supremum.