How do you find #1/(1+2^-1)xx1/(1+3^-1)xx1/(1+4^-1)xx...xx1/(1+n^-1)#?
3 Answers
Jan 20, 2016
Explanation:
Jan 20, 2016
#prod_(k=2)^n (1/(1+k^(-1))) = 2/(n+1)#
Explanation:
#1/(1+k^(-1)) = k/(k+1)#
So:
#prod_(k=2)^n (1/(1+k^(-1))) = 2/3 xx 3/4 xx ... xx n/(n+1) = 2/(n+1)#
Jan 20, 2016
Just a different style of writing the same thing as the others!
Explanation:
Write as:
Let the sum be
Numerator
Denominator
So we have: