Euler once sunu gosterdi: Harmonik seri
$H=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k}=\displaystyle\prod_{p\in \mathbb{P}}\dfrac{1}{1-\frac{1}{p}}$ olur.
Iki tarafin logaritmasini alalim..
$\ln H=\ln \Bigg(\displaystyle\prod_{p\in \mathbb{P}}\dfrac{1}{1-\frac{1}{p}}\Bigg)=-\ln\Big(1-\dfrac{1}{2}\Big)-\ln\Big(1-\dfrac{1}{3}\Big)-\ln\Big(1-\dfrac{1}{5}\Big)-\dots$ $(1)$
$\ln(1+x)$'in Taylor serisisini yazalim.
$\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dots$
$\ln (1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dots$
$-\ln (1-x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\dots$ bizim isimize bu yarar. $(1)$ deki sag taraftaki herbir terimi Taylor serisi olarak yazalim..
$\ln H=\dfrac{1}{2}+\dfrac{\big(\frac{1}{2}\big)^2}{2}+\dfrac{\big(\frac{1}{2}\big)^3}{3}+\dfrac{\big(\frac{1}{2}\big)^4}{4}+\dots$
$+\dfrac{1}{3}+\dfrac{\big(\frac{1}{3}\big)^2}{2}+\dfrac{\big(\frac{1}{3}\big)^3}{3}+\dfrac{\big(\frac{1}{3}\big)^4}{4}+\dots$
$+\dfrac{1}{5}+\dfrac{\big(\frac{1}{5}\big)^2}{2}+\dfrac{\big(\frac{1}{5}\big)^3}{3}+\dfrac{\big(\frac{1}{5}\big)^4}{4}+\dots$
$\vdots$
Alt alta toplarsak.
$\ln H=\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p}+\dfrac{1}{2}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^2}+\dfrac{1}{3}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^3}+\dfrac{1}{4}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^4}+\dots$
Integral Testinden sunu biliyoruz, $n\geq2$ icin,
$\displaystyle\sum_{k=2}^{\infty}\dfrac{1}{k^n}\leq\displaystyle\int_1^{\infty}\dfrac{1}{x^n}\,dx=\dfrac{1}{n-1}$
$\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^2}\leq\displaystyle\sum_{k=2}^{\infty}\dfrac{1}{k^2}\leq\displaystyle\int_1^{\infty}\dfrac{1}{x^2}\,dx=\dfrac{1}{2-1}=1$
$\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^3}\leq\dfrac{1}{3-1}=\dfrac{1}{2}$
$\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^4}\leq\dfrac{1}{4-1}=\dfrac{1}{3}$
$\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^5}\leq\dfrac{1}{5-1}=\dfrac{1}{5}$
$\vdots$
$\dfrac{1}{2}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^2}+\dfrac{1}{3}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^3}+\dfrac{1}{4}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^4}+\dots$
$\leq\dfrac{1}{2}1+\dfrac{1}{3}\dfrac{1}{2}+\dfrac{1}{4}\dfrac{1}{3}+\dfrac{1}{5}\dfrac{1}{4}+\dots$
$=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dots$
$\Bigg(1-\dfrac{1}{2}\Bigg)+\Bigg(\dfrac{1}{2}-\dfrac{1}{3}\Bigg)+\Bigg(\dfrac{1}{3}-\dfrac{1}{4}\Bigg)+\Bigg(\dfrac{1}{4}-\dfrac{1}{5}\Bigg)\dots=1$
Boylece
$\ln H=\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p}+\underbrace{\dfrac{1}{2}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^2}+\dfrac{1}{3}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^3}+\dfrac{1}{4}\displaystyle\sum_{p\in \mathbb{P}}\dfrac{1}{p^4}+\dots}_{\leq1}$
$H=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k}$ Harmonik serisi iraksaktir ve dolayisiyla $\ln H$ iraksaktir. Sol taraf iraksak oldugundan sag taraf da iraksak olur.