Question

Calculate, to four decimal places, the first eight terms of the recursive sequence. Does it appear to be convergent? If so, guess the value of the limit. Then assume the limit exists and determine its exact value. a1=1,an+1=61+ana_{1}=1, a_{n+1}=\frac{6}{1+a_{n}}

Solution

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The given recursive sequence is a1=1a_{1}=1 and

an+1=61+ana_{n+1}=\dfrac{6}{1+a_{n}}

for all nn. Let us calculate the eight terms of the sequence up to four decimal places

a1=1a_{1}=1

a2=61+a1=61+1=3a_{2}=\dfrac{6}{1+a_{1}}=\dfrac{6}{1+1}=3

a3=61+a2=61+3=1.5000a_{3}=\dfrac{6}{1+a_{2}}=\dfrac{6}{1+3}= 1.5000

a4=61+a3=61+1.5000=2.4000a_{4}=\dfrac{6}{1+a_{3}}=\dfrac{6}{1+1.5000}=2.4000

a5=61+a4=61+2.4000=1.7647a_{5}=\dfrac{6}{1+a_{4}}=\dfrac{6}{1+2.4000}= 1.7647

a6=61+a5=61+1.7647=2.1702a_{6}=\dfrac{6}{1+a_{5}}=\dfrac{6}{1+1.7647 }= 2.1702

a7=61+a6=61+2.1702=1.8926a_{7}=\dfrac{6}{1+a_{6}}=\dfrac{6}{1+2.1702}= 1.8926

a8=61+a7=61+1.8926=2.0742a_{8}=\dfrac{6}{1+a_{7}}=\dfrac{6}{1+1.8926}= 2.0742

The sequence seems to oscillate around 22 and gradually gets closer to 22. So the sequence seems to converge and our guess for the limit is 2.2.

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