Convergence of P-Series

Theorem
Let $p$ be a complex number.

If $\Re(p) > 1$:
 * $\displaystyle \sum_{n=1}^\infty n^{-p}$

converges absolutely.

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

Convergent Case
Let $p=x+iy$. Then


 * $\displaystyle \sum_{n=1}^\infty \left|{n^{-p}}\right| = \sum_{n=1}^\infty \frac 1 {\left|{n^x n^{iy}}\right|} = \sum_{n=1}^\infty \frac 1 {\left|{n^x e^{-iy \log (n)}}\right|} = \sum_{n=1}^\infty \frac 1 {\left|{n^x}\right| \left|{e^{-iy \log (n)}}\right|} = \sum_{n=1}^\infty\frac 1 {\left|{n^x}\right|}$

by Euler's Formula.

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

Then:


 * $\displaystyle \sum_{n=1}^\infty \frac 1 {n^x}$ converges if and only if $\displaystyle \int_1^\infty \frac{\mathrm dt}{t^x}$ converges, by the integral test.

But:


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

Since $x>1, 1-x < 0$ 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:


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

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.