TU Wien:Mathematik 1 VO (Panholzer)/Prüfung 2010-06-25

Beispiel 3

• Für $n=1$:

${\begin{aligned}\sum _{k=1}^{1}k*H_{1}&={\frac {1*(1+1)}{2}}*H_{k}-{\frac {1*(1-1)}{4}}\\1*{\frac {1}{1}}&={\frac {2}{2}}-{\frac {0}{4}}\\1&=1\end{aligned}}$

• Für $n=n+1$:

Beispiel 5

${\begin{aligned}{\frac {1}{k*(k+2)}}&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\\s_{n}&=\sum _{k=1}^{n}{\frac {1}{k*(k+2)}}=\sum _{k=1}^{n}{\frac {\frac {1}{2}}{k}}-\sum _{k=1}^{n}{\frac {\frac {1}{2}}{k+2}}\\&={\frac {\frac {1}{2}}{1}}+{\frac {\frac {1}{2}}{2}}+{\frac {\frac {1}{2}}{3}}+\cdots +{\frac {\frac {1}{2}}{n}}-{\frac {\frac {1}{2}}{3}}-\cdots -{\frac {\frac {1}{2}}{n}}-{\frac {\frac {1}{2}}{n+1}}-{\frac {\frac {1}{2}}{n+2}}\\&={\frac {\frac {1}{2}}{1}}+{\frac {\frac {1}{2}}{2}}-{\frac {\frac {1}{2}}{n+1}}-{\frac {\frac {1}{2}}{n+2}}\end{aligned}}$ ${\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k*(k+2)}}=\lim _{x\to \infty }sn=\lim _{x\to \infty }{\frac {\frac {1}{2}}{1}}+{\frac {\frac {1}{2}}{2}}-{\frac {\frac {1}{2}}{n+1}}-{\frac {\frac {1}{2}}{n+2}}={\frac {3}{4}}\end{aligned}}$