The question is related to the behavior of
Grandi's series,
i.e. the divergent infinite series •
S = 1 − 1 + 1 − 1 + 1 − 1 + · · · For even values of
n, the above finite series sums to 1; for odd values, it sums to 0. In other words, as
n takes the values of each of the non-negative
integers 0, 1, 2, 3, ... in turn, the series generates the
sequence {1, 0, 1, 0, ...}, representing the changing state of the lamp. The sequence does not
converge as
n tends to infinity, so neither does the infinite series. Another way of illustrating this problem is to rearrange the series: •
S = 1 − (1 − 1 + 1 − 1 + 1 − 1 + · · ·) The unending series in the parentheses is exactly the same as the original series
S. This means
S = 1 −
S which implies
S = 1⁄2. In fact, this manipulation can be rigorously justified: there are
generalized definitions for the sums of series that do assign Grandi's series the value 1⁄2. One of Thomson's objectives in his original 1954 paper is to differentiate supertasks from their series analogies. He writes of the lamp and Grandi's series, Later, he claims that even the divergence of a series does not provide information about its supertask: "The impossibility of a super-task does not depend at all on whether some vaguely-felt-to-be-associated arithmetical sequence is convergent or divergent." == See also ==