| 1 | 1 | |||
1− | < (1− | )n  | ||
| n+2 | n2+2n+1 |
| n | 1 | |||
Myślałem po lewej o 1− | , jednak nie odpowiada mi 1− | |||
| n2+2n+1 | n+2+1n |
| 1 | n | n | 1 | |||||
(1 − | )n > 1− | > 1 − | = 1 − | |||||
| n2+2n+1 | n2+2n+1 | n2+2n | n+2 |
| 1 | ||
x = − | , oczywiście x>−1 | |
| n2+2n+1 |
| n | 1 | |||
−xn = | = − | |||
| n2+2n+1 | n+2+1/n |
| 1 | 1 | 1 | 1 | |||||
n+2+1/n> n+2 więc | < | więc − | >− | |||||
| n+2+1/n | n+2 | n+2+1/n | n+2 |
| 1 | 1 | 1 | ||||
zatem (1− | )n>1− | >1− | ||||
| n2+2n+1 | n2+2+1/n | n+2 |