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Let $(x_n), n \ge 1,$ be a sequence nonnegative real numbers. If for any $n \ge 1$, $$|x_n - 4| < \frac{4n}{n^3+1},$$ then $(\sqrt{x_n})$ is convergent and find it's limit. Explain without any theorem.

Attempt: First of all, I want to prove that $x_n \to 4$ for $n \to \infty$. Let $\epsilon>0$ be given. By the Arcimedean Principle, we know that there exists a positive integer $N$ such that $\frac{4}{N} < \epsilon.$ Hence, for any positive integer $n$ with $n \ge N$, we have $$|x_n-4|<\frac{4n}{n^3+1} < \frac{4n}{n^3} = \frac{4}{n^2} < \frac{4}{n} \le \frac{4}{N} < \epsilon.$$ Therefore, $x_n \to 4$.

Now, I want to prove that $\sqrt{x_n} \to 2$. Since $x_n \to 4$, then for any $\epsilon > 0$, there exists a positive integer $N$ such that for all positive integer $n$ with $n \ge N$, we have $|x_n - 4|< 2\epsilon$. This equivalent to $$\frac{|x_n-4|}{2} < \epsilon. \tag{1}$$ Hence, for all positive integer $n$ with $n \ge N$, we have \begin{align*} |\sqrt{x_n}-2| &= \frac{|x_n-4|}{\sqrt{x_n}+2} \\ &\le \frac{|x_n-4|}{2} \\ &< \epsilon. \qquad \qquad \qquad \qquad (\text{by (1)}) \end{align*}

Therefore, $\sqrt{x_n} \to 2$.

Is it correct?

math404
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1 Answers1

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Per your question, your proof is correct. But you can save a step or two if you use the $|\sqrt{x_n} - 2| = \dfrac{|x_n - 4|}{\sqrt{x_n} + 2} \le \dfrac{|x_n - 4|}{2} < \dfrac{1}{2}\cdot \dfrac{4}{n} = \dfrac{2}{n}$. This is a bit quicker as you easily choose the $N$ for a given $\epsilon > 0$.

Wang YeFei
  • 6,390