#2 29. 11. 2007 08:07 — Editoval Marian (29. 11. 2007 12:21)
Re: Konvergence
Tzv. nutna podminka konvergence nekonecne ciselne rady 
se povazuje splneni podminky 
. Jeji nesplneni implikuje divergenci diskutovane rady. Dale polozim
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Nyni![$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left (\frac{n-1}{n+1}\right )^{n^2+4n+5}=\nl \hspace{64mm}=\left (1+\frac{-2}{n+1}\right )^{n^2+4n+5}=\nl \hspace{64mm}=\lim_{n\to\infty}\left [\left (1+\frac{-2}{n+1}\right )^{n+1}\right ]^{\frac{n^2+4n+5}{n+1}}=\nl \hspace{64mm}=\lim_{n\to\infty}\left [\underbrace{\left (1+\frac{-2}{n+1}\right )^{n+1}}_{\to e^{-2}}\right ]^{\overbrace{\frac{n+4+5/n}{1+1/n}}^{\to \frac{n+4}{1}=n+4}}=\nl \hspace{64mm}=\lim_{n\to\infty}\left (e^{-2}\right )^{n+4}=0. $](/mathtex/a8/a835d6d7de5be6a0f93ccda140580412.gif "kopírovat do textarea")
Proto nutna podminka konvergence nasi rady je splnena. Pro samotne vysetreni nasi rady je mozne pouzit hned nekolik kriterii. Nejjednodussi v tomto pripade je asi limitni odmocninove kriterium, kde se studuje hodnota limity ![$\lim_{n\to\infty}\sqrt[n]{a_n}$](/mathtex/f0/f07853de533063ecf75b9fdf176f8464.gif "kopírovat do textarea")
. V nasem pripade je ale 
pro vsechna cisla 
. Tudiz![$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\sqrt[n]{\left (\frac{n-1}{n+1}\right )^{n^2+4n+5}}\ =\nl \hspace{75mm}=\lim_{n\to\infty}\left [\left (1+\frac{-2}{n+1}\right )^{n^2+4n+5}\right ]^{1/n}=\nl \hspace{75mm}=\lim_{n\to\infty}\left (1+\frac{-2}{n+1}\right )^{\overbrace{\frac{n^2+4n+5}{n}}^{=n+4+5/n\to n+4}}=\nl \hspace{75mm}=\lim_{n\to\infty}\left (1+\frac{-2}{n+1}\right )^{n+4}=\cdots =e^{-2}<1$](/mathtex/aa/aa754b62e135a26533bc4ef934f65780.gif "kopírovat do textarea")
.
Studovana rada je tedy konvergentni.
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