Question #9cb16

1 Answer
Aug 22, 2016

Answer:

Any convergent sequence of real numbers will converge to a real number.

Explanation:

If we look at the question directly as asked, then the answer is yes, as the real numbers are a subset of the complex numbers. Then, any convergent sequence of reals will converge to a complex number.

Assuming, however, that the question is asking whether there can be a sequence of real numbers which converge to a number with an imaginary component, the answer is no.

A sequence #a_n# is said to converge to some limit #L# if for any #epsilon > 0# there exists an integer #N>0# such that #n>=N# implies #|a_n-L| < epsilon#. If we extend this to the complex numbers, then we can quickly see why the convergence in question cannot take place.

Suppose #a_n# is a sequence of real numbers, and #a+bi# is a complex number with #b!=0#. Then, for any #a_n#, we have

#|a_n-a+bi| = sqrt((a_n-a)^2+b^2) >= sqrt(b^2) = |b|#.

This shows directly that given any proposed limit with an imaginary component, we can demonstrate that the sequence does not converge to that limit by setting #epsilon = |b|#.