Kinia :

	1
Olicz sume: ∑ nad ∞, pod n=1. wyrazenie : ln(1−		)
	(n+1)2

Bogdan:

	1		(n + k − 1)(n + k + 1)
1 −		=
	(n + k)2		(n + k)(n + k)

log_a(x₁) + log_a(x₂) + log_a(x₃) + ... + log_a(x_k) = log_a(x₁*x₂*x₃*...*x_k) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

	n(n+2)		(n+1)(n+3)		(n+2)(n+4)
ln[ (		*		*		*... ]
	(n+1)(n+1)		(n+2)(n+2)		(n+3)(n+3)

co dostrzegasz?