How do you evaluate the sum represented by #sum_(n=1)^(8)1/(n+1)# ?

1 Answer
Oct 22, 2014

Begin by changing the denominators to #1+1, 2+1, 3+1# and so on to #8+1...#

Next add #1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9#

This requires a common denominator. If we multiply #9*8*7*5# we will get #2520#.

The #9# picks up multiples of the #3#, the and the #8# picks up multiples of the #2, 3 and 4#.

Now, multiply #1/2*1260/1260# giving #1260/2520#. Multiply #1/3*840/840# giving #840/2520#.

#1/4*630/630=630/2520, 1/5*504/504=504/2520, 1/6*420/420=420/2520, 1/7*360/360=360/2520, 1/8*315/315=315/2520 and 1/9*280/280=280/2520.#

Finally, add

#(1260+840+630+504+420+360+315+280)/2520# which = #4609/2520#.