A proof of the divergence of the harmonic series

Alexander Frolkin

8 August 2001

We start off with the following definition of the Riemann Zeta function:

$$\zeta(s) := \frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{{\rm e}^x - 1} {\rm d}x$$

We turn our attention to the integrand:

$$\frac{x^{s-1}}{{\rm e}^x - 1} = x^{s-1} \frac{1}{{\rm e}^x} ... ...}^{-x}} = {\rm e}^{-x} x^{s-1} \sum_{r=0}^\infty {\rm e}^{-rx}$$

Some manipulation of the series yields

$${\rm e}^{-x} x^{s-1} \sum_{r=0}^\infty {\rm e}^{-rx} = \sum_... ...e}^{-(r+1)x} x^{s-1} = \sum_{r=1}^\infty {\rm e}^{-rx} x^{s-1}$$

Plugging back into the Zeta function,

$$\zeta(s) = \frac{1}{\Gamma(s)}\int_0^\infty \sum_{r=1}^\inft... ...sum_{r=1}^\infty \int_0^\infty {\rm e}^{-rx} x^{s-1} {\rm d}x$$

Substitute $u=rx$ in the integral¹, so that

$$\int_0^\infty {\rm e}^{-rx} x^{s-1} {\rm d}x = \int_0^\inft... ...^\infty {\rm e}^{-u} u^{s-1} {\rm d}u = \frac{\Gamma(s)}{r^s}$$

Plugging back into the definition,

$$\zeta(s) = \frac{1}{\Gamma(s)} \sum_{r=1}^\infty \frac{\Gamma(s)}{r^s}$$

This leads to the more common definition of the Riemann Zeta function,

$$\zeta(s) = \sum_{r=1}^\infty \frac{1}{r^s}$$

If we put $s=1$, the function reduces to the harmonic series --

$$\zeta(1) = \sum_{r=1}^\infty \frac 1r$$

We can find the value of $\zeta(1)$ using the integral definition. So, we have

$$\zeta(1) = \frac{1}{\Gamma(1)} \int_0^\infty \frac{{\rm d}x}{{\rm e}^x - 1}$$

To evaluate the integral

$$\mathcal{I} := \int_0^\infty \frac{{\rm d}x}{{\rm e}^x - 1}$$

substitute

$$u = {\rm e}^x \mbox{ and so } {\rm d}x = \frac{{\rm d}u}{u}$$

Hence,

$$\mathcal{I} = \int_1^\infty \frac{{\rm d}u}{(u-1)u} = \int_1^\infty \frac{{\rm d}u}{u - 1} - \int_1^\infty \frac{{\rm d}u}{u}$$

Since these are improper integrals, in the Riemann sense, we need to take limits. Now,

$$\int_a^b \frac{{\rm d}u}{u-1} - \int_1^b \frac{{\rm d}u}{u} ... ...) - \ln(a-1) - \ln b = \ln\left(1 - \frac 1b\right) - \ln(a-1)$$

We have

$$\mathcal{I} = \lim_{b\to\infty, a\to 1^+} \ln\left(1 - \fra... ...ght) - \ln(a - 1) = \lim_{a \to 1^+} - \ln(a - 1) = -(-\infty)$$

Since

$$\Gamma(1) = \int_0^\infty {\rm e}^{-y} {\rm d}y = 1$$

We have

$$\zeta(1) = \mathcal{I} = \infty$$

Hence,

$$\sum_{r=1}^\infty \frac 1r = \infty$$

completing the proof that the harmonic series diverges.$\Box$