(n+1)!−n! | n!(n+1)−n!) | n!(n+1−1) | n | |||||
an= | = | = | = | |||||
(n+1)!+n! | n!(n+1)+n! | n!(n+1+1) | n+2 |
1 | ||
wiec lim to | ? a odp mam 2 | |
2 |
n*1 | ||
limn→∞ | =1 | |
n(1+2n) |
n+1 | ||
i monotoniczność an+1 = | ||
n+3 |
n+1 | n | (n+1)(n+2)−n(n+3) | 2 | |||||
an+1−an= | − | = | = | wiec jest | ||||
n+3 | n+2 | (n+3)(n+2) | (n+3)(n+2) |