Convergence of P-Series/Lemma

Theorem
Let $p = x + i y$ be a complex number where $x, y \in \R$ such that:
 * $x > 0$
 * $x \ne 1$

Then:
 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges $\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.

Proof
Let $p = x + i y$.

Then:

by Euler's Formula.

Now since $x > 0$, and all $n \ge 1$, all terms are positive and we may do away with the absolute values.

Then by the Integral Test:


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

First let $x \ne 1$:

Hence:


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




 * $\ds \int_1^{\to \infty} \frac {\d t} {t^x} + \dfrac {1^{1 - x} } {1 - x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} }$

and so:
 * $\ds \sum_{n \mathop = 1}^\infty \size {n^{-p} }$ converges


 * $\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.
 * $\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.

For $x = 1$:

which diverges.