Telescoping product

Calculus Level 2

A(x)=x+1x+1x+1 A(x)=x+\cfrac { 1 }{ x+\cfrac { 1 }{ x+\cfrac { 1 }{ \ddots } } } With A(x)A(x) defined as above, where the continued fraction goes on indefinitely, find the value of the infinite product 1A(1)×1+11A(1)×1+11+11A(1)×. \dfrac { 1 }{ A(1) } \times \dfrac { 1+\frac { 1 }{ 1 } }{ A(1) } \times \frac { 1+\frac { 1 }{ 1+\frac { 1 }{ 1 } } }{ A(1) } \times \cdots . If your answer can be expressed as a+bcb,\dfrac { a+\sqrt { b } }{ c\sqrt { b } }, where a,b,ca, b, c are positive integers and bb is square-free, give your answer as 100a+10b+c100a+10b+c.


Bonus: Can you give a closed formula for yA(y)×y+1yA(y)×y+1y+1yA(y)×\dfrac { y }{ A(y) } \times \dfrac { y+\frac { 1 }{ y } }{ A(y) } \times \dfrac { y+\frac { 1 }{ y+\frac { 1 }{ y } } }{ A(y) } \times \cdots when it converges?

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