In the previous installment, we determined that if in the endless fraction, , where P and S are positive integers such that S = P + 1, we consider the sequence of fractions formed by ending the fraction at successive occurrences of P or S and call it f (n), where n is the total number of occurrences of P and S in the fraction, then the endless fraction is equal to f (∞) = lim f (n) as n→∞ , and since when S = P + 1, we get the conjuction of two sequences with lim f (n) as n→∞ = P for n = an odd number and lim f (n) as n→∞ = 1 for n = an even number, the lim f (n) as n→∞ does not exist for P ≠ 1, so the fraction whose value equals that limit has no value, which gives us a mathematical justification for rejecting the algebraic solution to the problem of determining the value of an endless fraction satisfying the given conditions.
But we need not restrict ourselves to the given (or any) conditions. For any values of P and S, the value of the endless fraction is f (∞) = lim f (n) as n→∞ , so if we can find lim f (n) as n→∞ , we will have found the value of the fraction. To find lim f (n) as n→∞ , it may be useful to express f (n) in terms of P and S.
f (1) = P
f (2) =
f (3) = =
f (4) = =
f (5) = =
f (6) = =
f (7) = =
It is clear that this sequence can be defined recursively (see Puzzle 48) as f (1) = P , f (2) = , and for all n ≥ 3 , f (n) = . Then lim f (n) as n→∞ = lim as n→∞ . Let lim f (n) as n→∞ = L It follows that lim f (n – 2) as n→∞ = L , so L = , LS – L² = P and L² – SL + P = 0. This equation has the same structure and the same roots as the one resulting from the algebraic evaluation of the endless fraction. This m,eans that the solution of this equation yields equal roots if and only if the limit L exists. If the solution of the equation yields unequal roots, the limit we are seeking does not exist and we have our mathematical justification for rejecting both values.
In summary, an endless fraction of the type explored above is meaningless if the equation that results from the standard algebraic approach to evaluating it yields unequal roots. The fraction has a value if and only if the equation has equal roots.