## Precalculus: Mathematics for Calculus, 7th Edition

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