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

Question

Isabella Ackerman · Accepted Answer

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#.

Ezekiel Alexander · Answer

undefined To evaluate the sum $ \sum_{n=1}^{8} \frac{1}{n+1} $, you need to add up the terms of the series where $ n $ ranges from 1 to 8.

$$ \sum_{n=1}^{8} \frac{1}{n+1} = \frac{1}{1+1} + \frac{1}{2+1} + \frac{1}{3+1} + \frac{1}{4+1} + \frac{1}{5+1} + \frac{1}{6+1} + \frac{1}{7+1} + \frac{1}{8+1} $$

$$ = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} $$

$$ = 0.5 + 0.333 + 0.25 + 0.2 + 0.167 + 0.143 + 0.125 + 0.111 $$

$$ = 1.717 $$

So, the sum $ \sum_{n=1}^{8} \frac{1}{n+1} $ evaluates to approximately 1.717.