Convergence of P-Series

Theorem
Let $p$ be a complex number.

If $\Re \left({p}\right) > 1$, then the $p$-series:
 * $\displaystyle \sum_{n \mathop = 1}^\infty n^{-p}$

converges absolutely.

If $0 < \Re \left({p}\right) \le 1$, the series diverges.

Convergent Case
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 iff $\displaystyle \int_1^\infty \frac{\mathrm dt}{t^x}$ converges.

But:


 * $\displaystyle \int_1^{\to \infty} \frac{\mathrm d t}{t^x} = \left({\lim_{t \to \infty} \frac{t^{1-x}}{1-x}}\right) - \left({ \frac{1^{1-x}}{1-x} }\right)$

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


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

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:


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

Divergent Case
As proved above, the convergence of the $p$-series is dependent on the convergence of:


 * $\displaystyle \lim_{t \to\infty} \frac{t^{1-x}}{1-x}$

If $x = 1$, the series clearly diverges because of Division by Zero.

Suppose $0 < x < 1$.

Then:

Again, the result follows from the Integral Test.

Also see
The mapping $\displaystyle p \mapsto \sum_{n \mathop = 1}^\infty n^{-p}$ is well-known as the Riemann zeta function.