x | ||
limbn > | ||
x+1 |
1 | x | |||
a2= x2* | =x* | |||
x+1 | x+1 |
1 | x2 | x | ||||
a3= x3* | =x* | =x*( | )2 | |||
(x+1)2 | (x+1)2 | x+1 |
1 | x3 | x | ||||
a4=x4* | =x* | =x*( | )3 | |||
(x+1)3 | (x+1)3 | x+1 |
x | ||
an= x*( | )n−1 | |
x+1 |
x | x | x | ||||
bn=a1+a2+a3+a4+....+an=x+x* | +x*( | )2+x*( | )3+. | |||
x+1 | x+1 | x+1 |
x | ||
..+x*( | )n−1 | |
x+1 |
x | x | x | ||||
bn=x[1+ | +( | )2+(U{x}[x+1})3+.....( | )n−1] | |||
x+1 | x+1 | x+1 |
x | ||
q= | i |q|<1 to w nawiasie kwadratowym | |
x+1 |
1−(x/(x+1)n | x | |||
bn=x* | więc lin bn i n→∞= | |||
1−(x/(x+1) | 1−(x/(x+1) |
x | ||
| | |<1 | |
x+1 |
x | x | ||
> | co już nie problem | ||
1−(x/(x+1) | x+1 |