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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:

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

We turn our attention to the integrand:

\begin{displaymath}
\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}
\end{displaymath}

Some manipulation of the series yields

\begin{displaymath}
{\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}
\end{displaymath}

Plugging back into the Zeta function,

\begin{displaymath}
\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
\end{displaymath}

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

\begin{displaymath}
\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}
\end{displaymath}

Plugging back into the definition,

\begin{displaymath}
\zeta(s) = \frac{1}{\Gamma(s)} \sum_{r=1}^\infty \frac{\Gamma(s)}{r^s}
\end{displaymath}

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

\begin{displaymath}
\zeta(s) = \sum_{r=1}^\infty \frac{1}{r^s}
\end{displaymath}

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

\begin{displaymath}
\zeta(1) = \sum_{r=1}^\infty \frac 1r
\end{displaymath}

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

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

To evaluate the integral

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

substitute

\begin{displaymath}
u = {\rm e}^x \mbox{ and so } {\rm d}x = \frac{{\rm d}u}{u}
\end{displaymath}

Hence,

\begin{displaymath}
\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}
\end{displaymath}

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

\begin{displaymath}
\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)
\end{displaymath}

We have

\begin{displaymath}
\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)
\end{displaymath}

Since

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

We have

\begin{displaymath}
\zeta(1) = \mathcal{I} = \infty
\end{displaymath}

Hence,

\begin{displaymath}
\sum_{r=1}^\infty \frac 1r = \infty
\end{displaymath}

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


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Alexander Frolkin 2001-08-16