Question #280116

Check the convergence of the sequence defined by š‘¢š‘›+1 = (1 + 1/ š‘¢š‘› ) , š‘¢1 > 0. Note that this is the sequence associated with the continued fraction expansion of the Golden ratio. 


Expert's answer

Solution:

Given sequence, un+1=1+1un,u1>0u_{n+1}=1+\dfrac1{u_n},u_1>0

Let the sequence is convergent to ll.

∓l=1+1l⇒l2=l+1⇒l2āˆ’lāˆ’1=0⇒l=1±52∵u1>0∓l=1+52\therefore l=1+\dfrac 1l \\ \Rightarrow l^2=l+1 \\ \Rightarrow l^2-l-1=0 \\ \Rightarrow l=\dfrac{1\pm\sqrt5}{2} \\ \because u_1>0 \\\therefore l=\dfrac{1+\sqrt5}{2}

Hence, the given sequence is convergent to 1+52\dfrac{1+\sqrt5}{2} .


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