Oppg:
[tex]$$\sum\limits_{n = 1}^\infty {{{{{\left( {n + 2} \right)}^n}} \over {n!}}} $$[/tex]
Løsningsforslag:
[tex]$$L = {\lim }\limits_{n \to \infty } \frac{\frac{\left ( \left ( n+1 \right ) +2\right )^2}{n! \cdot \left ( n+1 \right )}} {\frac{\left ( n+2 \right )^2}{n!}} \cdot \frac{\frac{n!}{\left ( n+2 \right )^2}}{\frac{n!}{\left ( n+2 \right )^2}}$$[/tex]
[tex]$$der\;\left( {n + 1} \right)! = n! \cdot \left( {n + 1} \right)$$[/tex]
[tex]$$L = {\lim }\limits_{n \to \infty } {{{{\left( {n + 3} \right)}^2}} \over {\left( {n + 1} \right){{\left( {n + 2} \right)}^2}}}$$[/tex]
[tex]$$L = {\lim }\limits_{n \to \infty } {{{n^2} + 6n + 9} \over {{n^3} + 5{n^2} + 8n + 4}}$$[/tex]
Deler på den høyeste potensen:
[tex]$$L = {\lim }\limits_{n \to \infty } \frac{\frac{1}{n}+\frac{6}{n^2}+\frac{9}{n^3}}{1+\frac{5}{n}+\frac{8}{n^2}+\frac{4}{n^3}} = {{0 + 0 + 0} \over {1 + 0 + 0 + 0}} = 0 $$[/tex]
[tex]$$ \Rightarrow \; L \;<\; 1$$[/tex] rekka konvergerer.
