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

Finally, add

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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.

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