How do you find #1/(1+2^-1)xx1/(1+3^-1)xx1/(1+4^-1)xx...xx1/(1+n^-1)#?

Question

1436 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-20T12:25:14

#prod_(k=2)^n1/(1+k^(-1))=2/(n+1)#

#prod_(k=2)^n1/(1+k^(-1)) = prod_(k=2)^n(k*1)/(k(1+k^(-1)))#

#= prod_(k=2)^nk/(k+1)#

#=(2*3*4*...*n)/(3*4*5*...*(n+1))#

#=2/(n+1)#

Answer 2 · 2016-01-20T12:25:27

#prod_(k=2)^n (1/(1+k^(-1))) = 2/(n+1)#

#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)#

Answer 3 · 2016-01-20T15:22:45

#s=2/(n+1)#

Just a different style of writing the same thing as the others!

Write as:

Let the sum be #s#

#s=1/(1+1/2)xx1/(1+1/3)xx1/(1+1/4)xx ... xx 1/(1+1/n) #

#s=1/(3/2) xx1/(4/3)xx1/(5/4)xx ... xx1/((n+1)/n)#

#s=2/3xx3/4xx4/5xx ...xx n/(n+1)#

Numerator#-> n!#

Denominator#->((n+1)!)/2#

So we have:#color(white)(..)s= (2n!)/((n+1)!)#

#s=(2cancel(n!))/(cancel(n!)(n+1))#

#s=2/(n+1)#