Is it possible to define such an operator $\operatorname{\Gamma}$ that satisfies $ \lim_{n\to \infty} {\Gamma} (f(n))=\beta $?

Question

Let, the function $f:\mathbb N^+ \longrightarrow \mathbb R$ is given. Is it possible to define such an operator $\operatorname{\Gamma}$ that satisfies the following conditions:

For any $n\in\mathbb {N^+}$ we have

$$\operatorname{\Gamma} (f(n))=\alpha$$and

$$\displaystyle \lim_{n\to \infty} \operatorname{\Gamma} (f(n))=\beta $$

where $\alpha,\beta \in \mathbb R$ and $\alpha\neq \beta.$

Is this mathematically possible?

Please check Prove: If a sequence converges, then every subsequence converges to the same limit.. — John Omielan, Apr 05 '20 at 01:48
Good work! Yes, you are correct, it's not possible. I see there's now an answer, but I hope that other post helps you understand better why it's not possible. — John Omielan, Apr 05 '20 at 01:51

score 1 · Accepted Answer · answered Apr 05 '20 at 01:51

1

No, simply because $$\lim_{n\to\infty}\Gamma(f(n)) = \lim_{n\to\infty}\alpha = \alpha\neq\beta$$

answered Apr 05 '20 at 01:51

MPW

43,638

Wrong answer. ${}$ – Apr 07 '20 at 16:42
The answer is correct for the question as asked. Perhaps you stated the question incorrectly? – MPW Apr 07 '20 at 17:25
Yes, you are most likely right. I may not have asked the question correctly. I accept your answer. If I want to ask new question, I will ask for a new one. – Apr 07 '20 at 17:31

Is it possible to define such an operator $\operatorname{\Gamma}$ that satisfies $ \lim_{n\to \infty} {\Gamma} (f(n))=\beta $?

1 Answers1