Convergence of P-Series/Divergence if p between 0 and 1
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Theorem
Let $p$ be a complex number.
Let $0 < \map \Re p \le 1$.
Then the $p$-series:
- $\ds \sum_{n \mathop = 1}^\infty n^{-p}$
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.
$\Box$
Hence, the convergence of the $p$-series is dependent on the convergence of:
- $\ds \lim_{t \mathop \to \infty} \frac {t^{1 - x} } {1 - x}$
Suppose $0 < x < 1$.
Then:
\(\ds \lim_{t \mathop \to \infty} \frac {t^{1 - x} } {1 - x}\) | \(=\) | \(\ds \frac 1 {1 - x} \lim_{t \mathop \to \infty} t^{1 - x}\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds +\infty\) | Limit at Infinity of $x^n$ |
The special case of $x = 1$ is covered by Integral of Reciprocal is Divergent.
Hence the result from the Cauchy Integral Test.
$\blacksquare$