6
$\begingroup$

Tried the squeeze theorem but it didn't get me anywhere since:

$0 \leftarrow \frac{1}{n^{1-1/n}}=\frac{n^{1/n}}{n}\leq\frac{(1^n+2^n+...+n^n)^{1/n}}{n}\leq \ n^{1/n}\to 1$

None of the other tools we studied seem to work. Any tips? Thanks.

  • 0
    your expression is greater than ${(n^n)^{1 \over n} \over n} = 1$, so is bounded below by $1$ for all $n$. So the squeeze theorem applies as you were trying to do.2011-02-05

1 Answers 1

9

$\frac{\left(1^n+2^n+...+n^n\right)^{1/n}}{n} = \left(\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+...+1^n\right)^{1/n} \geq (1)^{1/n} = 1 \rightarrow 1$

The rest is as you suggest.

  • 0
    Sure, sorry, mistyped. Corrected.2011-02-05