Problem
ALG-B1-M05-P012 Invariant in a Fractional Recurrence
#12
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Let \(a_1=2\), \(a_{n+1}=\frac{2a_n}{a_n+2}\). Find \(\frac1{a_n}\).
Pass to reciprocals.
\(\frac1{a_{n+1}}=\frac{a_n+2}{2a_n}=\frac12+\frac1{a_n}\). Thus \(\frac1{a_n}=\frac12+\frac{n-1}{2}=\frac n2\). Hence \(a_n=\frac2n\).
The substitution turns the recurrence into an arithmetic progression.