Man untersuche die Folge (an)n≥1{\displaystyle (a_{n})_{n\geq 1}}auf Konvergenz und bestimmte gegebenenfalls den Grenzwert, indem man zwei geeignete Folgen (bn)n≥1{\displaystyle (b_{n})_{n\geq 1}}, (cn)n≥1{\displaystyle (c_{n})_{n\geq 1}}mit bn≤an≤cn{\displaystyle b_{n}\leq a_{n}\leq c_{n}}finde.
(a) an=1n2+1+1n2+2+...+1n2+n{\displaystyle a_{n}={\frac {1}{n^{2}+1}}+{\frac {1}{n^{2}+2}}+...+{\frac {1}{n^{2}+n}}}
(b) an=1n2+1+1n2+2+...+1n2+n{\displaystyle a_{n}={\frac {1}{\sqrt {n^{2}+1}}}+{\frac {1}{\sqrt {n^{2}+2}}}+...+{\frac {1}{\sqrt {n^{2}+n}}}}
Sandwich Theorem, Satz 4.22, Seite 161
bn=1n2+n+1n2+n+1n2+n+...=nn2+n=1n+1{\displaystyle 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}}}
limn→∞1n+1=0{\displaystyle \lim _{n\to \infty }{\frac {1}{n+1}}=0}
cn=1n2+1n2+1n2+...=nn2=1n{\displaystyle c_{n}={\frac {1}{n^{2}}}+{\frac {1}{n^{2}}}+{\frac {1}{n^{2}}}+...={\frac {n}{n^{2}}}={\frac {1}{n}}}
limn→∞1n=0{\displaystyle \lim _{n\to \infty }{\frac {1}{n}}=0}
bn=1n2+n+1n2+n+1n2+n+...=nn2+n=nn2∗(1+1n)=nn∗(1+1n)=1(1+1n){\displaystyle 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}})}}}}
limn→∞1(1+1n)=limn→∞1(1+0)=1{\displaystyle \lim _{n\to \infty }{\frac {1}{\sqrt {(1+{\frac {1}{n}})}}}=\lim _{n\to \infty }{\frac {1}{\sqrt {(1+0)}}}=1}
cn=1n2+1+1n2+1+1n2+1+...=nn2+1=nn2∗(1+1n2)=nn∗1+1n2=11+1n2{\displaystyle 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}}}}}}}
limn→∞11+1n2=limn→∞11+0=1{\displaystyle \lim _{n\to \infty }{\frac {1}{\sqrt {1+{\frac {1}{n^{2}}}}}}=\lim _{n\to \infty }{\frac {1}{\sqrt {1+0}}}=1}
