Answer
$\dfrac{29}{20}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
