Question

Question #9cb16

Answer 1

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|`.