Question #80860

2 Answers

Notice that the given be written as follows

#(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n+1))#

regroup and we have

#(1/1)+(-1/2+1/2)+(-1/3+1/3)+(-1/n+1/n)-1/(n+1)#

The sums inside the parentheses are zero hence we left with

#(1/1-1/((n+1)))=(n+1-1)/(n+1)=n/(n+1)#

Sep 3, 2016

See below.

Explanation:

For #n=1# is true for

#1/(1 xx 2) = 1/2#

Supposing that is true for #n#

#sum_(k=1)^n1/(k(k+1)) = n/(n+1)#

then it must be true for #n+1#

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

then by mathematical induction, the affirmation is true.