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Vy�et�eme, zda konverguje nebo diverguje �ada:

$$\sum\limits^\infty_{n=1} {n^n \over n!}$$.

\subheading{�e�en�}
Pou�ijeme limitn�ho d 'Alembertova krit�ria:

$${a_{n+1}\over a_n} = {{(n+1)^{(n+1)}\over (n+1)!}\over{n^n \over n!}} = {(n+1)(n+1)^n \over n^n (n+1)} = \left(1+{1 \over n}\right)^n$$

Limita tohoto v�razu je rovna $e > 1$ a dan� �ada je tedy divergentn�.









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