Answer
$S_{n}=-\log \left( n+1\right) $
$S_{1}=-\log2\approx -0.301$
$S_{2}=-\log 3\approx -0.477$
$S_{3}=-\log 4\approx -0.602$
$S_{4}=-\log 5\approx -0.699 $
Work Step by Step
$a_{n}=\log \left( \dfrac {n}{n+1}\right) =\log \left( n\right) -\log \left( n+1\right) $
$S_{1}=a_{1}=\log 1-\log 2=-\log2\approx -0.301$
$S_{2}=a_{1}+a_{2}=\left( \log 1-\log 2\right) +\left( \log 2-\log 3\right) =\log 1-\log 3=-\log 3\approx -0.477$
$S_{3}=a_{1}+a_{2}+a_{3}=\log 1-\log 2+\log 2-\log 3+\log 3-\log 4=\log 1-\log 4=-\log 4\approx -0.602$
$S_{4}=a_{1}+a_{2}+a_{3}+a_{4}=\log 1-\log 2+\log 2-\log 3+\log 3-\log 4+\log 4-\log 5=\log 1-\log 5=-\log 5\approx -0.699 $
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$S_{n}=-\log \left( n+1\right) $
