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Let $\{x_n\}$ and $\{y_n\}$ be positive sequences. Assume $\lim_{n\to\infty} \frac{x_n}{y_n} = 0$. I have to prove if the claim $\lim_{n\to\infty} x_n \div \frac{x_n + y_n}{2} = 0$ is always true, always false, or sometimes true and sometimes false depending on the sequences.

My attempt: \begin{align*} x_n + y_n &\geq y_n \\ 0 < \frac{1}{x_n + y_n} &\leq \frac{1}{y_n} \\ 0 < \frac{x_n}{x_n + y_n} &\leq \frac{x_n}{y_n} \end{align*} Then by Squeeze Theorem, $\lim_{n\to\infty} \frac{x_n}{x_n + y_n} = 0 \implies \lim_{n\to\infty} \frac{2x_n}{x_n + y_n} = 0$ which is what we wanted. Could you check if my proof makes sense. Is there any cases of sequences of x_n and y_n such that this statement is not true?

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    $\begingroup$ Your proof is fine ! $\endgroup$
    – Fred
    Mar 4, 2021 at 11:03
  • $\begingroup$ @Fred: I have one trivial question: can I still use squeeze theorem for $0 < \frac{x_n}{x_n + y_n} \leq \frac{x_n}{y_n}$ if I don't have $0 \leq $ rather than $0 <$? $\endgroup$
    – user894272
    Mar 4, 2021 at 11:08
  • $\begingroup$ @abrakadabra_01 Yes, you can still use squeeze theorem in this case, the type of inequality doesn't matter for when applying the squeeze theorem for problems of the form $a_n \leq b_n \leq c_n$ as long as $\lim_{n\to\infty}a_n = \lim_{n\to\infty}c_n$. $\endgroup$
    – spaceman
    Mar 4, 2021 at 11:27

2 Answers 2

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Alternative approach

$\frac{y_n}{x_n} \to \infty \implies$
$\frac{1}{2} \times \frac{x_n + y_n}{x_n} = \frac{1}{2} \times \left[1 + \frac{y_n}{x_n}\right] \to \infty.$

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$$\frac{x_n}{x_n+y_n\over 2}=2\frac{x_n \over y_n}{{x_n\over y_n}+1}$$

Therefore $$\lim_{n\to \infty}\frac{x_n}{x_n+y_n\over 2}=2 \frac{0}{{0}+{1}}=0 $$

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