Answer
sum = $0+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{5}}$
Work Step by Step
$a_{j}=\sqrt{\frac{j-1}{j+1}}$
$j=1,\qquad a_{1}=\sqrt{\frac{1-1}{1+1}}=0$
$j=2,\qquad a_{2}=\sqrt{\frac{2-1}{2+1}}=\sqrt{\frac{1}{3}}$
$j=3, \qquad a_{3}=\sqrt{\frac{3-1}{3+1}}=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}$
$j=4, \qquad a_{3}=\sqrt{\frac{4-1}{4+1}}=\sqrt{\frac{3}{5}}$
sum = $0+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{5}}$
