Benutzer:Mogoh/sandbox4

Aufgabe 18

e)

$a_{n}=\prod _{k=2}^{n}\left(1-{\frac {1}{k^{2}}}\right)$

$\prod _{k=2}^{n}\left(1-{\frac {1}{k^{2}}}\right)=\prod _{k=2}^{n}\left({\frac {k^{2}-1}{k^{2}}}\right)$

= \prod_{k=2}^n \left(\frac{(k-1) \cdot (k+1)}{k \cdot k} \right)

= \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2 \cdot 4}{3 \cdot 3} \cdot \frac{3 \cdot 5}{ 4 \cdot 4} ... \cdot \frac{(n-1) (n+1)}{n \cdot n}

= \frac{1}{2} \cdot \frac{n+1}{n}

\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \prod_{k=2}^n \left(1 - \frac{1}{k^2} \right) = \lim_{n \rightarrow \infty} \frac{1}{2} \cdot \frac{n+1}{n} = \frac{1}{2}

f)

a_n = \frac{n^n}{3^n n!}

\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{3^{n+1} (n+1)!} \cdot \frac{3^nn!}{n^n} = \frac{1}{3} \frac{(n+1)^n}{n!} \frac{n!}{n^n}

= \frac{1}{3} \left( 1+ \frac{1}{n} \right) ^n < \frac{1}{3}e <1

\Rightarrow 0 \le a_n für n \in \mathbb{N} , a_n monoton fallend

\Rightarrow a_n ist konvergent

\frac{a_{n+1}}{a_n} < \frac{e}{3} \Leftrightarrow a_{n+1} < \frac{e}{3} a_n

d.h. a_2 < \frac{e}{3} a_1 a_3 < \frac{e}{3} a_2 < \frac{e}{3} \frac{e}{3} a_1 = \left( \frac{e}{3} \right)^2 a_1 a_4 < \left( \frac{e}{3} \right)^3 \frac{1}{3} a_{n+1} < \left( \frac{e}{3} \right)^n \frac{1}{3}

(n \rightarrow \infty )

\Rightarrow \lim_{n \rightarrow \infty} a_n = 0

g)

a_n = \left( 1-\frac{1}{n^2} \righ)^n

• (i) 1-\frac{1}{n^2} < 1 \Rightarrow \left( 1-\frac{1}{n^2} \righ)^n = a_n < 1
• (ii) \left( 1-\frac{1}{n^2} \righ)^n = \left( 1 + \left( -\frac{1}{n^2} \right) \righ)^n > 1 - \frac[n}{n^2} = 1 - \frac{1}{n}

(Bernullli Ungleichung

-\frac{1}{n^2} > -1 für n \ge 2

\Rightarrow 1 -\frac{1}{n} < a_n < 1 für alle n \ge 2

\Rightarrow \lim_{n \righarrow \infty} a_n = 1

Beispiel zu Teleskopreihe

\sum_{n=1}^\infty \frac{1}{\left( n - \frac{1}{3} \right)\left( n + \frac{8}{3} \right)}

\frac{1}{\left( n - \frac{1}{3} \right)\left( n + \frac{8}{3} \right)} = \frac{A}{\left( n - \frac{1}{3} \right)} + \frac{1}{\left( n + \frac{8}{3} \right)}

1 = A (n+ \frac{8}{3}) + B \left(n-\frac{1}{3} \right)

n = \frac{8}{3}

1 = B \left(- \frac{8}{3} -\frac{1}{3} \right) = -B3 \Leftrightarrow B = - \frac{1}{3}

n = \frac{1}{3}

1 = A \left(\frac{1}{3} +\frac{8}{3} \right) = 3A \Rightarrow A = \frac{1}{3} \Leftrightarrow B = - \frac{1}{3}

Probe/Beweis

\frac{\frac{1}{3}}{n-\frac{1}{3}} - \frac{\frac{1}{3}}{n + \frac{8}{3}} = \frac{1}{3} \left( \frac{n + \frac{8}{3} - \left( n-\frac{1}{3} \right)}{\left( n-\frac{1}{3} \right) \left( n+\frac{8}{3} \right)} \right)

= \frac{1}{\left( n-\frac{1}{3} \right) \left( n+\frac{8}{3} \right)}

\sum_{n=1}^\infty \frac{1}{\left( n-\frac{1}{3} \right) \left( n+\frac{8}{3} \right)} = \sum_{n=1}^\infty \left( \frac{1/3}{n-1/3} - \frac{1/3}{n-1/3+3} \right) = \frac{1/3}{n-1/3} + \frac{1/3}{2-1/3} + \frac{1/3}{3-1/3} = \frac{33}{40}

Kriterien

Wurzelkriterium

\limsup_{n \rightarrow \infty} \sqrt[n]{\vert a_n \vert} < 1 \Rightarrwo \sum_{n = 1}^\infty a_n kovergiert

Quoteinetenkriterium

\limsup_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \Rightarrwo \sum_{n = 1}^\infty a_n kovergiert

\sum_{n= 1}^\ifnty \frac{n+1}{2^n}

WK

\sqrt[n]\frac{n+1}{2^n} = \frac{\sqrt[n]{n+1}}{2} \xrightarrow[]{n \rightarrow \infty} \frac{1}{2}

\Rightarrow \limsup_{n \rightarrow \infty} \sqrt[n]{\frac{n+1}{2^n}} = \frac{1}{2} \Rightarrow \sum_{n=1}^\infty \frac{n+1}{2^n} < \infty

QK

\frac{n+2}{n^{n+1}} \frac{2^n}{n+1} = \frac{n+2}{n+1} \frac{1}{2} \xrightarrow[]{n \righarrow \infty} \frac{1}{2}

\Rightarrow \limsup \left| \frac{a_{n+1}}{a_n} \right| \frac{1}{2}

\sum_{n=1}^\infty \frac{n!}{(2n)!}