## College Algebra (11th Edition)

$\dfrac{29}{20}$
$\bf{\text{Solution Outline:}}$ To evaluate the given summation expression, $\displaystyle\sum_{i=1}^5 (i+1)^{-1} ,$ substitute $i$ with the values from $1$ to $5$ and then simplify the expression. $\bf{\text{Solution Details:}}$ Substituting $i$ with the numbers from $1$ to $5 ,$ the given expression evaluates to \begin{array}{l}\require{cancel} (1+1)^{-1}+(2+1)^{-1}+(3+1)^{-1}+(4+1)^{-1}+(5+1)^{-1} \\\\= (2)^{-1}+(3)^{-1}+(4)^{-1}+(5)^{-1}+(6)^{-1} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{2^1}+\dfrac{1}{3^1}+\dfrac{1}{4^1}+\dfrac{1}{5^1}+\dfrac{1}{6^1} \\\\= \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6} \\\\= \dfrac{30}{60}+\dfrac{20}{60}+\dfrac{15}{60}+\dfrac{12}{60}+\dfrac{10}{60} \\\\= \dfrac{87}{60} \\\\= \dfrac{\cancel3\cdot29}{\cancel3\cdot20} \\\\= \dfrac{29}{20} .\end{array}