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

Beispiel 3

• Für $n=1$:

$\begin{array}{cc}\sum _{k=1}^{1}k\ast H_1& =\frac{1\ast (1+1)}{2}\ast H_k-\frac{1\ast (1-1)}{4}\\ 1\ast \frac{1}{1}& =\frac{2}{2}-\frac{0}{4}\\ 1& =1\end{array}$

• Für $n=n+1$:

$\begin{array}{cc}\sum _{k=1}^{n+1}k\ast H_k& =\frac{(n+1)\ast ((n+1)+1)}{2}\ast H_{n+1}-\frac{(n+1)\ast ((n+1)-1)}{4}\\ \sum _{k=1}^{n}k\ast H_k+\sum _{k=n+1}^{n+1}k\ast H_k& =\frac{(n+1)\ast (n+2)}{2}\ast H_{n+1}-\frac{(n+1)\ast (n)}{4}\\ \sum _{k=1}^{n}k\ast H_k+(n+1)\ast H_{n+1}& =\frac{(n+1)\ast (n+2)}{2}\ast H_{n+1}-\frac{(n+1)\ast (n)}{4}\\ \frac{n\ast (n+1)}{2}\ast H_n-\frac{n\ast (n-1)}{4}+(n+1)\ast H_{n+1}& =\frac{(n+1)\ast (n+2)}{2}\ast H_{n+1}-\frac{(n+1)\ast (n)}{4}\\ (n\ast (n+1))\ast H_n-\frac{n\ast (n-1)}{2}+2\ast (n+1)\ast (H_n+\frac{1}{n+1})& =((n+1)\ast (n+2))\ast (H_n+\frac{1}{n+1})-\frac{n^2+n}{2}\\ (n^2+n)\ast H_n-\frac{n\ast (n-1)}{2}+2\ast (n+1)\ast (H_n+\frac{1}{n+1})& =(n^2+3n+2)\ast (H_n+\frac{1}{n+1})-\frac{n^2+n}{2}\\ n^2{H}_n+n{H}_n-\frac{n^2}{2}+\frac{n}{2}+2\ast (n+1)\ast H_n+\frac{2\ast (n+1)}{(n+1)}& =n^2{H}_n+3n{H}_n+2H_n+\frac{(n+1)\ast (n+2)}{n+1}-\frac{n^2}{2}-\frac{n}{2}\\ n^2{H}_n+n{H}_n-\frac{n^2}{2}+\frac{n}{2}+2n{H}_n+2H_n+2& =n^2{H}_n+3n{H}_n+2H_n+n+2-\frac{n^2}{2}-\frac{n}{2}\\ 3n{H}_n+\frac{n}{2}& =3n{H}_n+\frac{n}{2}\end{array}$

Beispiel 5

$\begin{array}{cc}\frac{1}{k\ast (k+2)}& =\frac{\frac{1}{2}}{k}-\frac{\frac{1}{2}}{k+2}\\ s_n& =\sum _{k=1}^n\frac{1}{k\ast (k+2)}=\sum _{k=1}^n\frac{\frac{1}{2}}{k}-\sum _{k=1}^n\frac{\frac{1}{2}}{k+2}\\ & =\frac{\frac{1}{2}}{1}+\frac{\frac{1}{2}}{2}+\frac{\frac{1}{2}}{3}+\cdots +\frac{\frac{1}{2}}{n}-\frac{\frac{1}{2}}{3}-\cdots -\frac{\frac{1}{2}}{n}-\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{2}}{n+2}\\ & =\frac{\frac{1}{2}}{1}+\frac{\frac{1}{2}}{2}-\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{2}}{n+2}\end{array}$ $\begin{array}{c}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\ast (k+2)}=\underset{x\to \mathrm{\infty }}{lim}sn=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{2}}{1}+\frac{\frac{1}{2}}{2}-\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{2}}{n+2}=\frac{3}{4}\end{array}$