Question

cum demonstram marginirea sirului x=1/3+2/21+3/91+...+n/(n^4+n^2+1)

Answer 1

   
[tex]\displaystyle\\ x_n = \frac{1}{3} + \frac{2}{21}+ \frac{3}{92}+\cdots + \frac{n}{n^4+n^2+1}\\\\ x_n \ \textgreater \ 0 ~~\texttt{dearece toti termenii sirului sunt pozitivi} \\ \Longrightarrow~~x_n ~~\texttt{este marginit inferior.}\\\\ \texttt{Pentru a demonstra ca sirul este marginit superior, }\\ \texttt{va trebui sa gasim un sir majorant cu limita cunoscuta.}\\\\ y_n= \frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\cdots+\frac{1}{n^2} \\\\ y_n~~\texttt{este un sir majorant pentru }x_n ~~~\texttt{deoarece:} [/tex]


[tex]\displaystyle\bf\\ \frac{1}{2}\ \textgreater \ \frac{1}{3};~\frac{1}{4}\ \textgreater \ \frac{3}{92};~\frac{1}{8}\ \textgreater \ \frac{3}{92};~\cdots \frac{1}{n^2}\ \textgreater \ \frac{n}{n^4+n^2+1}\\\\ \lim_{n\to\infty}y_n=\lim_{n\to\infty}\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\cdots+\frac{1}{n^2}=1\\\\ \Longrightarrow ~x_n \ \textless \ 1 \Longrightarrow~x_n~~\texttt{este marginit superior.}\\\\ \texttt{Concluzie:}\\\\ 0\ \textless \ x_n\ \textless \ 1 \\\\ \Longrightarrow ~x_n ~~\texttt{este marginit inferior si superior.}\\\\ \Longrightarrow ~\boxed{x_n ~~\texttt{\bf este marginit}}[/tex]