| 3n−1 | ||
an= | ||
| n |
| 3(n+1)−1 | 3n+2 | |||
an+1= | = | |||
| n+1 | n+1 |
| 3n+2 | 3n−1 | n(3n+2)−(3n−1)(n+1) | ||||
an+1−an= | − | = | = | |||
| n+1 | n | n(n+1) |
| 3n2+2n−3n2−2n+1 | 1 | |||
= | = | , gdzie n∊N+ | ||
| n(n+1) | n(n+1) |
| 3n − 1 | 1 | |||
an = | = 3 − | < 3 (a nie 2) | ||
| n | n |
an + 1 − an > 0 ⇒ an rośnie
| 1 | ||
Tak, z dołu przez 2 z góry przez 3. Granicą jest 3, bo przy n→∞ | → 0 | |
| n |