Question #9cb16

Aug 22, 2016

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 $n \ge N$ 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 $a + b i$ is a complex number with $b \ne 0$. Then, for any ${a}_{n}$, we have

$| {a}_{n} - a + b i | = \sqrt{{\left({a}_{n} - a\right)}^{2} + {b}^{2}} \ge \sqrt{{b}^{2}} = | b |$.

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 $\epsilon = | b |$.