\(\begin{align} A_1 & = \sqrt{3} \\ A_2 & = \sqrt{A_1} \\ A_3 & = \sqrt{A_2} \\ & \vdots \\ A_{n+1} &= \sqrt{A_n} \end{align}\)

What is the value of the infinite product \(A_1 \times A_2 \times A_3\times \cdots \, ?\)

Guest Jul 6, 2020

#1**0 **

s=1;a=1;listfor(i, 1, 25,s=(2#(3)) * 2#s

OUTPUT =(1.732050808, 2.279507057, 2.615056629, 2.800923042, 2.898753029, 2.948942029, 2.974361459, 2.987153223, 2.99356972, 2.996783135, 2.998391136, 2.99919546, 2.999597703, 2.999798845, 2.999899421, 2.99994971, 2.999974855, 2.999987427, 2.999993714, 2.999996857, 2.999998428, 2.999999214, 2.999999607, 2.999999804, 2.999999902) =**It converges to 3**

Guest Jul 6, 2020

#2**0 **

Let A_{1}·A_{2}·A_{3}· ... = X

Square both sides:

( A_{1}·A_{2}·A_{3}· ... )^{2} = X^{2}

(A_{1})^{2}·[ (A_{2})^{2}·(A_{3})^{2}·(A_{4})^{2}· ... ] = X^{2}

But: (A_{1}) = 3 (A_{2})^{2} = A_{1} (A_{3})^{2} = A_{2} (A_{4})^{2} = A_{3}

So: (A_{1})^{2}·[ (A_{2})^{2}·(A_{3})^{2}·(A_{4})^{2}· ... ] = 3·[ A_{1}·A_{2}·A_{3}· ... ] = X^{2}

and: A_{1}·A_{2}·A_{3}· ... = X ---> = 3·[ X ] = X^{2} ---> 3 = X

So, the product is **3**.

geno3141 Jul 6, 2020