The discussion centers on proving that the sequence sin(nπ/2) does not converge to 0. Participants demonstrate that while sin(nπ/2) alternates between 0, 1, and -1, establishing non-convergence to 0 requires using the non-existence of limit definition with epsilon. One participant suggests defining the sequence as (-1)^n for odd n to show non-convergence to ±1, while questioning how to represent the sequence's behavior towards 0. The conversation emphasizes the need for a general term equivalent to (-1)^n for demonstrating non-convergence to 0.
PREREQUISITESStudents and educators in mathematics, particularly those studying calculus and sequences, as well as anyone interested in understanding convergence and limit definitions in trigonometric contexts.
peripatein said:I have to demonstrate it using the non-existence of limit definition. Hence, using epsilon etc.
peripatein said:I have to demonstrate it using the non-existence of limit definition. Hence, using epsilon etc.
peripatein said:In order to prove that it does not converge to +-1, I simply defined it as (-1)^n and then showed that for any even n (or odd n, in the case of limit being +1) |(-1)^n - 1| >= epsilon (taking epsilon=1).
But what would be the general term, the equivalent of (-1)^n, for 0? I mean, I'd still have to prove that the sequence does not converge to 0 in order for my proof to be complete.
peripatein said:It is true and is equal to (-1)^n for all odd positions, i.e. n=2k+1.