∏j=1n+1(1+1/(n+1+j))
Indexverschiebung:
∏j=2n+2(1+1/(n+j))
= ∏j=1n+2(1+1/(n+j)) / ( 1+1/(n+1) )
= (1+1/(2n+1)∗(1+1/(2n+2)∗(∏j=1n(1+1/(n+j))/(1+1/(n+1))
=(1+1/(2n+1)∗(1+1/(2n+2)∗(2−1/(n+1))/(1+1/(n+1))
=( 2n+3)/(n+2)
= ( 2n+4)/(n+2) - 1/(n+2)
= 2- 1 / (n+2) . Bingo !