ISI2015-MMA-55

Question

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

• A. $\underset{n \to \infty}{\lim} a_{2n}$ and $\underset{n \to \infty}{\lim} a_{2n+2}$ exists
• B. $\underset{n \to \infty}{\lim} a_{2n}$ and $\underset{n \to \infty}{\lim} a_{2n+1}$ exist
• C. $\underset{n \to \infty}{\lim} a_{2n}, \underset{n \to \infty}{\lim} a_{2n+1}$ and $\underset{n \to \infty}{\lim} a_{3n}$ exist
• D. none of the above

Comments

Yes, you are correct. The sequence $(-1)^n$ doesn’t have a single limit point (if it had a single limit point, then it would have been convergent), but it has two limit points, i.e. limit of $a_n$ exists but it is not unique.

 

Moreover, the questions ask about the existence and not the convergence. Further, the theorem we both agree upon is the corollary of the theorem I quoted in my last comment.

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Okay , I will think it again ,

But as we know The limit of a sequence is unique then it means its subsequence also converges to the same limit point .

It means according to you if limit of a2n and limit a2n -1 exists then the limit of an exist ?

 But it contradicting the fact that limit limit of (-1)^n doesnot exist.

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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.

Answer 1

answer is D as all subsequence converges to the same limit let us call it l=m=n= k