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. $a_{1}=1, a_{n+1}=\frac{6}{1+a_{n}}$

Solution

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

$a_{n+1}=\dfrac{6}{1+a_{n}}$

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

$a_{1}=1$

$a_{2}=\dfrac{6}{1+a_{1}}=\dfrac{6}{1+1}=3$

$a_{3}=\dfrac{6}{1+a_{2}}=\dfrac{6}{1+3}= 1.5000$

$a_{4}=\dfrac{6}{1+a_{3}}=\dfrac{6}{1+1.5000}=2.4000$

$a_{5}=\dfrac{6}{1+a_{4}}=\dfrac{6}{1+2.4000}= 1.7647$

$a_{6}=\dfrac{6}{1+a_{5}}=\dfrac{6}{1+1.7647 }= 2.1702$

$a_{7}=\dfrac{6}{1+a_{6}}=\dfrac{6}{1+2.1702}= 1.8926$

$a_{8}=\dfrac{6}{1+a_{7}}=\dfrac{6}{1+1.8926}= 2.0742$

The sequence seems to oscillate around $2$ and gradually gets closer to $2$. So the sequence seems to converge and our guess for the limit is $2.$

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