Convergence of P-Series/Divergence if p between 0 and 1

Theorem

Let $p$ be a complex number.

Let $0 < \map \Re p \le 1$.

Then the $p$-series:

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

diverges.

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.

$\blacksquare$