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To show: $\lim_{n\rightarrow\infty} {\int_{1}^{n}\frac{1}{x^{n}} dx} = 0.$

I argued that for every $x>1$: $\lim_{n\rightarrow\infty} \frac{1}{x^{n}} = 0.$

However, is this proof rigorous?

Thanks for your advice.

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    @Mariano: I realize that you are probably right. However, rather than simply saying that "no, the argument is not rigorous," I figured that at least they could look up [Dominated Convergence](http://en.wikipedia.org/wiki/Dominated_convergence_theorem) and learn something about it.2011-10-11

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To be more explicit about the hints in the comments, you can just evaluate the integral in the usual way. $ \lim_{n \rightarrow \infty} \int_1^n \frac{1}{x^n} dx = \lim_{n \rightarrow \infty} \left[\frac{-1}{(n-1)x^{n-1}}\right]_1^n = \lim_{n \rightarrow \infty} \left[ \frac{-1}{(n-1)n^{n-1}} - \frac{-1}{(n-1)} \right]. $ Now that the integral is out of the way, all that remains is to show the limit above is zero.