Oppgave 1

Vi har at H_{n + 1} = H_n + 1/(n + 1) eller H_n = H_{n + 1} − 1/(n + 1).

Dermed:

(n − 1) · n! · (2H_n − 3)  +  2 · [ n! · (2H_n − 3) + n! ] =

((n − 1)n! +2n!) (2H_n − 3) + 2n! = (n + 1)! · (2H_n − 3) + 2n! =

(n + 1)! · (2H_{n + 1} − 3 − 2/(n + 1)) + 2n! = (n + 1)! · (2H_{n + 1} − 3) − 2n! + 2n! =

(n + 1)! · (2H_{n + 1} − 3)