# Prove that if $\lim_{n\to\infty} \frac{x_n}{y_n} = 0$, then $\lim_{n\to\infty} x_n \div \frac{x_n + y_n}{2} = 0$

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?

• Your proof is fine !
– Fred
Mar 4, 2021 at 11:03
• @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 $0 <$?
– user894272
Mar 4, 2021 at 11:08
• @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 $\lim_{n\to\infty}a_n = \lim_{n\to\infty}c_n$. Mar 4, 2021 at 11:27

$$\frac{y_n}{x_n} \to \infty \implies$$
$$\frac{1}{2} \times \frac{x_n + y_n}{x_n} = \frac{1}{2} \times \left[1 + \frac{y_n}{x_n}\right] \to \infty.$$
$$\frac{x_n}{x_n+y_n\over 2}=2\frac{x_n \over y_n}{{x_n\over y_n}+1}$$
Therefore $$\lim_{n\to \infty}\frac{x_n}{x_n+y_n\over 2}=2 \frac{0}{{0}+{1}}=0$$