TITLE: Computing $ \lim_{n\rightarrow\infty} (\sqrt[n]{1}+\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)^n $
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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 $e^{\log 2007!}=2007!$