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

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Theorem

Let $p$ be a complex number.

Let $\map \Re p > 1$.

Then the $p$-series:

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

converges absolutely.


Proof

Lemma

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


Since $x > 1$ it follows that $1 - x < 0$.

Thus $P^{1 - x} \to 0$ as $P \to \infty$.

Setting $x - 1 = \delta >0$, this limit is:

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

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:

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

$\blacksquare$


Sources


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