Limit of a sequence. Let $ { \{a_n \}} $ be a sequence of real numbers. Then $ \lim_{n \to \infty} a_n $ exists if and only if

Question

Let $ { \{a_n \}} $ be a sequence of real numbers. Then $ \lim_{n \to \infty} a_n $ exists if and only if

(A) $\lim_{n \to \infty} a_{2n}$ and $\lim_{n \to \infty} a_{2n+2}$ exists

(B) $\lim_{n \to \infty} a_{2n}$ and $\lim_{n \to \infty} a_{2n+1}$ exists

(C) $\lim_{n \to \infty} a_{2n}$ , $\lim_{n \to \infty} a_{2n+1}$ and $\lim_{n \to \infty} a_{3n}$ exists

(D) none of the above

So there is IFF condition. So for one of the option to be true the vice versa has to be true. I've considered many sequences like sequence of even numbers or $\{-1^n\} $ sequence. But still not getting any satisfactory answer. Would really appreciate if someone can help me understand the intuition.

PS: I've asked this question before. But has been deleted by the Community bot as it had 0 votes and 30 days older. I'm really looking for an answer. Hope this time someone can answer.

Answer 1

The correct answer is (C), not (D).

Assume that $L_0=\lim_{n\to\infty}a_{2n}$, $L_1=\lim_{n\to\infty}a_{2n+1}$, and $L_2=\lim_{n\to\infty}a_{3n}$ exist. The sequence $\langle a_{6n}:n\in\Bbb N\rangle$ is a subsequence of both $\langle a_{2n}:n\in\Bbb N\rangle$ and $\langle a_{3n}:n\in\Bbb N\rangle$, so it converges to both $L_0$ and $L_2$, which must therefore be equal. The sequence $\langle a_{6n+3}:n\in\Bbb N\rangle$ is a subsequence of both $\langle a_{2n+1}:n\in\Bbb N\rangle$ and $\langle a_{3n}:n\in\Bbb N\rangle$, so it converges to both $L_1$ and $L_2$, which must therefore be equal. In other words, $L_0=L_1=L_2$, and I’ll simply call it $L$ henceforth.

Let $\epsilon>0$. The sequence $\langle a_{2n}:n\in\Bbb N\rangle$ converges to $L$ so there is an $m_0\in\Bbb N$ such that $|a_{2n}-L|<\epsilon$ for all $n\ge m_0$. Similarly, there is an $m_1\in\Bbb N$ such that $|a_{2n+1}-L|<\epsilon$ for all $n\ge m_1$. Let $m=\max\{2m_0,2m_1+1\}$; if $n\ge m$ is even, then $\frac{m}2\ge m_0$, so $|a_n-L|<\epsilon$, and if $n$ is odd, then $\frac{n-1}2\ge m_1$, and again $|a_n-L|<\epsilon$. Thus, $|a_n-L|<\epsilon$ for all $n\ge m$, and since $\epsilon>0$ was arbitrary, we’ve shown that $\langle a_n:n\in\Bbb N\rangle$ converges to $L$.

Answer 2

By hint 1 the "only if"-statement is true in A, B and C. Since we do not know, that the limits are the same, all the "if"-statements are false, so statement D is correct.

Observations: 1. Even if the both limits in A coincide, we cannot conclude that lim a(n) converges. Counterexample: a(n)=1 if n is odd and a(n)=2 if n is even. Here both a(2n) and a(2n+2) have the same limit =2. But a(n) does not converge since there are two different accumulation points. 2. By hint 2, B and C are true if the all limits coincide.

Answer 3

For C, suppose $\lim a_{2n} = p$, $\lim a_{2n+1} = q$, $\lim a_{3n} = r$.

Considering $a_{6n}$, $p = r$.

Considering $a_{6n+3} =a_{2(3n+1)+1}$, $q = r$.

Therefore $p = q = r$.

Since $a_{2n}$ and $a_{2n+1}$ include all integers, $\lim a_n$ exists.

So C is true.