Question #9cb16

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Question #9cb16

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Answer 1 · 2016-08-22T09:43:17

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

Suppose $a_{n}$ $a_n$ is a sequence of real numbers, and $a + b i$ $a+bi$ is a complex number with $b \neq 0$ $b!=0$ . Then, for any $a_{n}$ $a_n$ , we have

$| a_{n} - a + b i | = \sqrt{{(a_{n} - a)}_{n}^{2} + b^{2}} \geq \sqrt{b^{2}} = | b |$ $|a_n-a+bi| = sqrt((a_n-a)^2+b^2) >= sqrt(b^2) = |b|$ .

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 $ε = | b |$ $epsilon = |b|$ .

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Question #9cb16

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