# Prove that $\displaystyle\lim_{n \rightarrow \infty}x_n \leq \displaystyle\lim_{n \rightarrow \infty}y_n.$

Suppose $(x_n)$ and $y_n$ are two convergent sequences such that infinitely many of the terms satisfy $x_n \leq y_n$. Prove that $\displaystyle\lim_{n \rightarrow \infty}x_n \leq \displaystyle\lim_{n \rightarrow \infty}y_n.$

My attempt: Since $x_n \leq y_n$ holds for infinitely many $n$, by choosing subsequence, we can assume that $x_n \leq y_n$ holds for all $n$. We can also assume that $(x_n)$ and $(y_n)$ are convergent. Therefore, $\displaystyle\lim_{n \rightarrow \infty} x_n \leq \displaystyle\lim_{n \rightarrow \infty} y_n.$

Is my attempt correct?

• Using argument by contradiction. – xpaul Apr 21 '17 at 12:57
• Your attempt looks pretty good indeed, yet instead of "We can also assume...", you could more accurately write "We know that ..." . Also, a sequence converges iff any subsequence converges to same limit as the sequence – DonAntonio Apr 21 '17 at 13:02
• Your attempt is quasi-circular, you just traded "infinitely many" for "all", but the main claim remains unchanged. – Yves Daoust Apr 21 '17 at 13:28
• You are wrong; one can be less than the other for infinitely many terms but still not be for all terms. – Jacob Wakem May 10 '17 at 20:23

• suppose that $\displaystyle\lim_{n \rightarrow \infty}x_n > \displaystyle\lim_{n \rightarrow \infty}y_n$;
• you can then find an $N$ such that $x_n > y_n$ for all $n > N$;
• but then $x_n \le y_n$ can only hold for at most $N$ and thus finitely many $n$.
• $x_n>y_n$ for infinitely many $n$ doesn't contradict that $x_n\le y_n$ for infinitely many $n$. However, what you proved does suffice: $x_n>y_n$ for all $n\ge N$, so $x_n\le y_n$ for at most $N<\infty$ values of $n$, namely $n\in\{1,2,\dots,N\}$. – Cody Johnson May 30 '18 at 18:44