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quasar987
Oct16-04, 07:13 PM
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}

quasar987
Oct16-04, 10:45 PM
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. :smile: