The proof that $\sum_{n=1}^{\infty} \frac{1}{n^k} $converges for all $k>1$.

Question

I saw this step being used in another proof. We have been shown that this is true of $k=2$, but not this general form.

Answer 1

Could this be done using the integral test? Using the integral \begin{equation} \int_1^\infty n^{-k} dn \end{equation} the series will be convergent if the integral is a finite value. If you evaluate the integral \begin{align} \int_1^\infty n^{-k} dn = \lim_{x \to \infty}( \frac{n^{1-k}}{1-k}\Big|_1^x) = \lim_{x \to \infty} \frac{x^{1-k}}{1-k} - \frac{1}{1-k} \end{align} You can clearly see that the limit is finite if $k > 1$(the limit is actually 0 in this case), otherwise the limit is not finite. So it follows that the series is convergent if $k > 1$.