| 2n−1 | ||
Obliczyć sumę szeregu ∑ | ||
| 2n |
| 2n | 1 | |||
Rozbiłam to na dwa ułamki ∑ | − | Wtedy suma drugiego ułamka jest równa 1, bo | ||
| 2n | 2n |
| 2n−1 | 1 | 2n−1 | ||||
S=∑(n=1 do ∞) | = | +∑(n=2 do ∞) | = | |||
| 2n | 2 | 2n |
| 1 | 2*(n+1)−1 | |||
= | +∑(n=1 do ∞) | = | ||
| 2 | 2n+1 |
| 1 | 1 | 2*n−1+2 | ||||
= | + | ∑(n=1 do ∞) | = | |||
| 2 | 2 | 2n |
| 1 | 1 | 2*n−1 | 1 | |||||
= | + | ∑(n=1 do ∞) | +∑(n=1 do∞)( | )n= | ||||
| 2 | 2 | 2n | 2 |
| 1 | 1 | 1 | ||||
= | + | S+∑(n=1 do∞)( | )n⇔ | |||
| 2 | 2 | 2 |
| 1 | 1 |
| ||||||||||||
S= | + | ⇔ | ||||||||||||
| 2 | 2 |
|
| 1 | 3 | ||
S= | |||
| 2 | 2 |