Convergence of P-Series

Theorem
If $$\Re(p) > 1 \ $$,

$$\sum_{n=1}^\infty n^{-p} \ $$

converges absolutely.

Proof
Let $$p=x+iy \ $$. Then

$$\sum_{n=1}^\infty \left|{n^{-p}}\right| = \sum_{n=1}^\infty \frac{1}{ \left|{  n^x n^{iy}  }\right|  } = \sum_{n=1}^\infty \frac{1}{\left|{  n^x e^{-iy \log (n)}  }\right|}  =  \sum_{n=1}^\infty \frac{1}{\left|{  n^x  }\right|   \left|{  e^{-iy \log (n)}  }\right|} =\sum_{n=1}^\infty\frac{1}{\left|{n^x}\right|} \  $$

by Euler's Formula. Now since $$x > 1 \ $$, and all $$n \geq 1 \ $$, all terms are positive and we may do away with the absolute values. Then

$$\sum_{n=1}^\infty \frac{1}{n^x} \ $$ converges if and only if $$\int_1^\infty \frac{dt}{t^x} \ $$ converges, by the integral test. But

$$\int_1^\infty \frac{dt}{t^x} = \left({ \lim_{t\to\infty} \frac{t^{1-x}}{1-x} }\right)   -\left({ \frac{1^{1-x}}{1-x} }\right) \ $$

Since $$x>1, 1-x < 0 \ $$ and so setting $$x-1 = \delta >0 \ $$, this limit is

$$ -\frac{1}{\delta} \lim_{t\to\infty} \frac{1}{ t^\delta} = 0 \ $$

hence the integral is just $$\frac{1}{1-x} \ $$ (that is, convergent) and so the sum converges as well. Since the terms of the sum were positive everywhere, it is absolutely convergent and hence so is

$$\sum_{n=1}^\infty n^{-p} \ $$