kb mehr für heute
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@ -195,20 +195,20 @@ $
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= &lim_(n->infinity) (2)/((n+1)) = 0 < 1 => "S ist konvergent" checkmark
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= &lim_(n->infinity) (2)/((n+1)) = 0 < 1 => "S ist konvergent" checkmark
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$
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$
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#pagebreak()
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=== Funktionen
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=== Funktionen
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 35]
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$ f: D --> W\ v |-> w = f(w) $
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$ f^: D --> W\ v |-> w = f(w) $
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==== Accumulation Point
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#quote()[
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#quote()[
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Does the series
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A value $a$ is called accumulation point of $D$ if there is a (non-constant)
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sequence $a_n$ in $D$ which converges to $a$.
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] @Vorlesung[1, P. 40]
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$ S = sum^(infinity)_(n=0) (2^n)/((n+1)^n) $
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- Ist dann nicht jeder Punkt (in Mengen mit mehr als einem Element) ein
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Sammelpunkt? Folgen können doch belibig definiert werden?
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converge?
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]
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$
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&lim_(n->infinity) root(n, abs((2^n)/((n+1)^n))) "(Wurzelkriterium)"\
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= &lim_(n->infinity) root(n, abs(((2)/((n+1)))^n)) \
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= &lim_(n->infinity) (2)/((n+1)) = 0 < 1 => "S ist konvergent" checkmark
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$
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