TU Wien:Analysis UE (diverse)/Übungen WS19/Beispiel 7

Man untersuche die Folge $(a_{n})_{n\geq 1}$auf Konvergenz und bestimmte gegebenenfalls den Grenzwert, indem man zwei geeignete Folgen $(b_{n})_{n\geq 1}$, $(c_{n})_{n\geq 1}$mit $b_{n}\leq a_{n}\leq c_{n}$finde.

(a) $a_{n}={\frac {1}{n^{2}+1}}+{\frac {1}{n^{2}+2}}+...+{\frac {1}{n^{2}+n}}$

(b) $a_{n}={\frac {1}{\sqrt {n^{2}+1}}}+{\frac {1}{\sqrt {n^{2}+2}}}+...+{\frac {1}{\sqrt {n^{2}+n}}}$

Hilfreiches

Sandwich Theorem, Satz 4.22, Seite 161

Lösungsvorschlag von Bisco

(a)

$b_{n}={\frac {1}{n^{2}+n}}+{\frac {1}{n^{2}+n}}+{\frac {1}{n^{2}+n}}+...={\frac {n}{n^{2}+n}}={\frac {1}{n+1}}$

$\lim _{n\to \infty }{\frac {1}{n+1}}=0$

$c_{n}={\frac {1}{n^{2}}}+{\frac {1}{n^{2}}}+{\frac {1}{n^{2}}}+...={\frac {n}{n^{2}}}={\frac {1}{n}}$

$\lim _{n\to \infty }{\frac {1}{n}}=0$

(b)

$b_{n}={\frac {1}{\sqrt {n^{2}+n}}}+{\frac {1}{\sqrt {n^{2}+n}}}+{\frac {1}{\sqrt {n^{2}+n}}}+...={\frac {n}{\sqrt {n^{2}+n}}}={\frac {n}{\sqrt {n^{2}*(1+{\frac {1}{n}})}}}={\frac {n}{n*{\sqrt {(1+{\frac {1}{n}})}}}}={\frac {1}{\sqrt {(1+{\frac {1}{n}})}}}$

$\lim _{n\to \infty }{\frac {1}{\sqrt {(1+{\frac {1}{n}})}}}=\lim _{n\to \infty }{\frac {1}{\sqrt {(1+0)}}}=1$

$c_{n}={\frac {1}{\sqrt {n^{2}+1}}}+{\frac {1}{\sqrt {n^{2}+1}}}+{\frac {1}{\sqrt {n^{2}+1}}}+...={\frac {n}{\sqrt {n^{2}+1}}}={\frac {n}{\sqrt {n^{2}*(1+{\frac {1}{n^{2}}})}}}={\frac {n}{n*{\sqrt {1+{\frac {1}{n^{2}}}}}}}={\frac {1}{\sqrt {1+{\frac {1}{n^{2}}}}}}$

$\lim _{n\to \infty }{\frac {1}{\sqrt {1+{\frac {1}{n^{2}}}}}}=\lim _{n\to \infty }{\frac {1}{\sqrt {1+0}}}=1$