Question

Calculate the successive convergents to
$\frac{1}{2+}\frac{1}{2+}\frac{1}{3+}\frac{1}{1+}\frac{1}{4+}\frac{1}{2+}\frac{1}{6}$

Answer 1

To Calculate the successive convergents to

$\frac{1}{2+}\frac{1}{2+}\frac{1}{3+}\frac{1}{1+}\frac{1}{4+}\frac{1}{2+}\frac{1}{6}$

Let the continued fraction be denoted by

$a_1+\frac{1}{a_2+}\frac{1}{a_3+}\frac{1}{a_4+}...;$

Then the first three convergents are

$\frac{1}{a_1},\frac{a_2}{(a_1{a}_2+1)},\frac{a_3{a}_2+1}{a_3(a_1{a}_2+1)+a_1};$

Here ,

$a_1=2,a_2=2,a_3=3,a_4=1,a_5=4,a_6=2,a_7=6$

Then , the successive convergent can be calculated as

$\frac{1}{2},\frac{2}{4+1},\frac{7}{3(4+1)+2},...$

$\frac{1}{2},\frac{2}{5},\frac{7}{15},\frac{9}{22},\frac{43}{105},\frac{95}{232},\frac{613}{1497}....$