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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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Why didn't you just evaluate the integral first before taking limits? – J. M. Oct 9 '11 at 13:29
It certainly isn't rigorous. J. M.'s suggestion is a good one, though. – Dylan Moreland Oct 9 '11 at 13:30
First of all: don't forget that the domain of integration also depends on $n$. Also: pointwise convergence of the integrand to zero isn't sufficient for the integral to vanish. Another example: put $f_n(x) := 1/(x+n)$ for $x > 1.$ Clearly, $\forall x, \; \lim_n f_n(x) = 0$, but $\lim_n \int_1^n f_n(x)$ has a finite, non-zero value (find it!). – Gerben Oct 9 '11 at 13:56
For $n>1$, note that $\int_1^n\frac{1}{x^n}\;\mathrm{d}x\le\int_1^\infty\frac{1}{x^n}\;\mathrm{d}x$. Then you can use Dominated Convergence to get that $\lim_{n\to\infty}\int_1^\infty\frac{1}{x^n}\;\mathrm{d}x=0$ by the pointwise convergence $\lim_{n\rightarrow\infty} \frac{1}{x^{n}} = 0$. – robjohn Oct 9 '11 at 14:16
The proposed argument is not rigorous. With some work on inequalities, it could be made rigorous. Or else you could use heavy machinery. However, with the given function, the problem can be solved using standard tools of first-year calculus, nothing fancy. Calculate! – André Nicolas Oct 9 '11 at 15:58
up vote 5 down vote accepted

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.

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