| log n | log n | |||
∑ | domyślam się że będzie ≤ | ale nie wiem jak dalej. | ||
| n2 √n | n2 |
| |||||||||||
∑ | |||||||||||
| 5n |
| log(n) | 1 | 1 | |||
≤ | a szereg ∑ | jest zbieżny (bo 3/2>1) | |||
| n2√n | n3/2 | n3/2 |
| 1 | |||||||||||
≤ | ||||||||||||
| 5n | 5n |
| log n | ||
∑ | nie jest przypadkiem rozbiezny? | |
| n2 |
| n2n | (n+1)2n+2 | |||
∑ | po pierwszych obliczeniach wyszło mi | |||
| (2n)! | (2n+2)(2n+1)*n2n |
| n2n | ||
∑ | ||
| (2n)! |
| (n+1)2n+2(2n)! | (n+1)2n+2 | ||
= | = | ||
| (2n+2)!n2n | (2n+2)(2n+1)n2n |
| e2 | ||||||||||||
= | → | ||||||||||||
| (2n+2)(2n+1) | 4 |