[tex]\lim \, (n+1)! \left( e-\sum_{\nu=1}^{n}\frac{1}{\nu !} \right) = 1[/tex] Докажете го.
Означаваме:
[tex]l_1 = \lim \, n \left((n+1)! \left( e-\sum_{\nu=1}^{n}\frac{1}{\nu !} \right)-1\right)[/tex]
[tex]l_2 = n\left( n \left((n+1)! \left( e-\sum_{\nu=1}^{n}\frac{1}{\nu !} \right)-1\right)-l_1\right)[/tex]
[tex]l_3 = n\left(n\left( n \left((n+1)! \left( e-\sum_{\nu=1}^{n}\frac{1}{\nu !} \right)-1\right)-l_1\right) - l_2 \right)[/tex]
...
и т.н.
Намерете [tex]\lim \,l_n[/tex]

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