# Computing $\lim_{n\rightarrow\infty} (\sqrt[n]{1}+\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)^n$

I'm trying to find:

$$\lim_{n\rightarrow\infty} (\sqrt[n]{1}+\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)^n$$

(Problem from CMJ)

We have:

$$k^{1/n}=1+\frac{\ln k}{n}+O(1/n^2)$$

$$\left( \sum_{k=1}^{2007}k^{1/n}-2006 \right)^n= \left(1+\frac{1}{n}\sum_{k=1}^{2007}\ln k+O(1/n^2) \right)^n \sim_{n\rightarrow\infty} \exp \left( \sum_{k=1}^{2007} \ln k \right)(=2007!)$$

I'm quite sure of my result but I could not check it numerically.

Do you agree with this limit?

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Yes. Your limit looks fine. You can simplify it further into $$\exp \left( \sum_{k=1}^{2007} \log (k) \right) = \exp \left( \log \left(\prod_{k=1}^{2007} k \right) \right) = \exp \left( \log (2007!) \right) = 2007!$$

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OK, thank you Marvis! – Chon Nov 18 '12 at 22:24

Take the log of your expression and replace $x=\frac{1}{n}$, you get the expression:

$$\frac{\log\left((\sum_{k=1}^{2007} k^x) - 2006\right)}{x}$$

First, show the numerator approaches zero as $x\to 0$. Then apply L'Hopital's rule, yielding a limit:

$$\sum_{k=1}^{2007} \log k = \log 2007!$$

That is the log of your limit, so your limit is $e^{\log 2007!}=2007!$

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