Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1

Theorem
Let $p$ be a complex number. Let $\map \Re p > 1$.

Then the $p$-series:
 * $\displaystyle \sum_{n \mathop = 1}^\infty n^{-p}$

converges absolutely.

Lemma
Let $p = x + i y$.

Then:

by Euler's Formula.

Now since $x > 1$, and all $n \ge 1$, all terms are positive and we may do away with the absolute values.

Then by the Integral Test:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges $\displaystyle \int_1^\infty \frac {\d t} {t^x}$ converges.

But:


 * $\displaystyle \int_1^{\to \infty} \frac {\d t} {t^x} = \paren {\lim_{t \mathop \to \infty} \frac {t^{1 - x} } {1 - x} } - \paren {\frac{1^{1 - x} } {1 - x} }$

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


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

hence the integral is just $\dfrac 1 {1 - x}$ (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:


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