# Computing the limit $\lim \limits_{n \rightarrow \infty} \int_1^n \frac{1}{x^n} dx$

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?

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
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.