Prove that $\sum_{n=1} 1/n^p$ converges uniformly for any $p > 1$

Question

Can someone please give me a formal proof that $\sum_{n=1} 1/n^p$ converges uniformly for any $p > 1$? By the $p$-test, we know that such a series converges, but that doesn't speak to the uniform convergence of the series.

Answer 1

The easiest way to prove your statement is using Cauchy's integral test. The integral $$\int_1^\infty\frac{1}{x^p}dx$$ exists for any $p>1$, meaning that the sum $$\sum_{n=1}^\infty\frac{1}{n^p}$$ also converges.

The integral is rather simple to calculate, since it equals

$$\int_1^\infty\frac1{x^p}dx = \lim_{b\to\infty}\int_1^b\frac1{x^p}dx = \lim_{b\to\infty}- \left.\frac{1}{x^{p-1}}\right|_1^b = \lim_{b\to\infty}\left(1 -\frac{1}{b^{p-1}}\right) = 1$$

Answer 2

If $p\leq0$, $a_p=\frac{1}{n^p}$ it is not infinitesimal, if $p>0$ the general term is decreasing $$ n^p<(n+1)^p $$ and so the criterion of condensation tell us that $$ \sum_{k=1}^\infty2^k\frac{1}{(2^k)^p}=\sum_{k=1}^\infty\frac{1}{(2^k)^{p-1}}=\sum_{k=1}^\infty\frac{1}{(2^{p-1})^{k}} $$ that is the geometrical series with general term $1/2^{p-1}$ and now you can easily finish.