# Find $\lim_{n\to \infty}\frac{(1^n+2^n+…+n^n)^{1/n}}{n}$

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.

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put that 1/n inside the upper paranthesis – user1708 Feb 5 '11 at 15:06
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. – Zarrax Feb 5 '11 at 15:54

## 1 Answer

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

The rest is as you suggest.

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shouldn't that be $\left((\frac{1}{n})^n+(\frac{2}{n})^n+...+1^n\right)^{1/n}$? – user6661 Feb 5 '11 at 15:19
Sure, sorry, mistyped. Corrected. – snow Feb 5 '11 at 15:22