Ana1[Bearbeiten]

Aufgabe 28[Bearbeiten]

\sum_{n,k \in \mathbb N \times \mathbb N}2^{-n+1} 3^{-2k-1}

= \sum_{k = 1}^\infty \sum_{n= 1}^\infty 2^{-n+1} 3^{-2k-1}

= \sum_{k = 1}^\infty 3^{-2k-1} \sum_{n= 1}^\infty 2^{-n+1}

= \frac{1}{3} \cdot 2 \sum_{k=1}^\infty \left( \frac{1}{9}^k \right) \sum_{n=1}^\infty \left( \frac{1}{2}^n \right)

= \frac{2}{3} \cdot \left( \frac{1}{1-\frac{1}{9}} - 1 \right) \left( \frac{1}{1-\frac{1}{2}} - 1 \right)

= \frac{3}{16}

\sum_{n,k \in \mathbb N \times \mathbb N} \frac{1}{(k^2+n^2)^s}

= \sum_{j=2}^\infty \sum_{k+n = j} \frac{1}{(k^2+n^2)^s}

Falls j = k+n , dann gilt \frac{1}{2} j^2 \le k^2 + n^2 \le j^2

\le \sum_{j=2}^\infty \frac{j-1}{\left( \frac{1}{2}j^2 \right)^s}

\le \sum_{j=2}^\infty s^s \frac{j}{j^{2s}}

= \sum_{j=2}^\infty 2^s \frac{1}{j^{2s-1}} < \infty

\ge \sum _{j=2}^\infty \frac{j-1}{j^{2s}}

\ge \sum _{j=2}^\infty \frac{1}{2} \frac{j}{j^{2s}}

> \infty

\sum_{n,k \in \mathbb N \times \mathbb N} {n \choose k} a^n b^k :

\sum_{k = 1}^\infty \sum_{n= 1}^\infty \left| {n \choose k} a^n b^k \right|

Es gilt : {n \choose k } = 0 , falls k > n

= \sum_{k = 1}^\infty \sum_{n= 1}^n {n \choose k} \vert a \vert^n \vert b \vert^k

= \sum_{k = 1}^\infty \vert a \vert^n \left( \sum_{n= 1}^n {n \choose k} \vert b \vert^k 1^{n-k} - 1 \right)

= \sum_{k = 1}^\infty \vert a \vert^n \left[ (1+ \vert b \vert^n -1 \left]

hier fehlt etwas ....