Request of suggestion for improving an answer about evaluation of $\lim_{n\rightarrow\infty} \frac{n\cdot 2^n }{3^n}$

Question

My question is related to the following OP

How can I evaluate $\lim_{n\rightarrow\infty} \frac{n\cdot 2^n }{3^n}$?

The answer is $0$. But how? I've searched some links but all of them just quite write the answer and no one shows the procedure.

In order to give an answer to the above mentioned OP, I've presumed that the asker was aware about the following basic facts:

for $0<a<1 \quad a^n \to 0$

for $a>1 \quad a^n \to +\infty$

and that his/her doubt was about the way to handling in a simply manner the indeterminate form $\frac{\infty}{\infty}$ or that is the same $\infty \cdot 0$.

My answer was the following

$$\frac{n\cdot 2^n }{3^n}\stackrel{\text{definitively}}<\frac{\left(\frac{e}2\right)^n\cdot 2^n }{3^n}=\frac{e^n}{3^n}=\left(\frac{e}3\right)^n\to 0$$

which is basically an application of the squeeze theorem and that make use of the following basic fact:

$$\left(\frac{e}2\right)^n\stackrel{\text{definitively}}>n$$

Despite all my efforts to explain these basic concepts, the answer has been finally deleted by moderators without any good explanation.

Do you have any idea how to improve the answer in such way that it was more clear and if possible not eligible for deletion.

EDIT NOTE

After the discussion here with Daniel Fischer I've clarified that the widely used and more correct terms for definitively is eventually.

The original post was finally revised as follow:

Note that eventually, notably for $n>6$, the following result holds

$$n<\left(\frac{e}{2}\right)^n$$

thus for squeeze theorem

$$\frac{n\cdot 2^n }{3^n}<\frac{\left(\frac{e}2\right)^n\cdot 2^n }{3^n}=\frac{e^n}{3^n}=\left(\frac{e}3\right)^n\to0$$

I think this is a good compromise that takes into account all the positive and negative comments, remarks and feedback received.

Answer 1

Under the assumption that you are genuinely looking for input, here are the things that, in my opinion, are problematic with your original answer, which reads:

$$\frac{n\cdot 2^n }{3^n}\stackrel{\text{definitively}}<\frac{\left(\frac{e}2\right)^n\cdot 2^n }{3^n}=\frac{e^n}{3^n}=\left(\frac{e}3\right)^n\to0$$

That major problem with this answer is that it does nothing to help the student who might actually ask the question as it was asked. The answer is going to going zooming over the head of the asker. In particular:

1. You have not explained the notation $$ \stackrel{\text{definitively}}< .$$ I don't claim to have seen every notation under the sun, but I have been working in mathematics for a while (both as a student and instructor), and have never seen that notation before. If I have never seen that notation before, it seems reasonable that someone taking their first calculus class has never seen it, either. The answer would be improved by eliminating that notation and explaining why $n < \left( \frac{\mathrm{e}}{2} \right)^n $ for sufficiently large $n$.
2. It may not be immediately obvious to an elementary student that the inequality $n < \left( \frac{\mathrm{e}}{2} \right)^n$ holds for sufficiently large $n$. A couple of words explaining this inequality would be an improvement (basically, don't assume that something that is obvious to you is obvious).
3. As you stated in your question here, you are invoking the squeeze theorem. It would be helpful to the elementary student if you at least used the words "squeeze theorem" in your answer.

4. The way that the squeeze theorem is typically stated (in the context of sequential limits), it reads something like the following:

   Suppose that $a_n < b_n < c_n $ for all $n$ and that $\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = L$. Then $\lim_{n\to\infty} b_n = L$.

   To the elementary student who doesn't know how to approach the original limit in the first place, the lack of a lower bound might not be clear. Adding a $0 < $ on the left-hand side of the original inequality would help, i.e. start with $$ 0 < \frac{n\cdot 2^n}{3^n}\dotsb $$ This is a minor thing to change, but it would drastically improve the pedagogical value of your answer.

5. The $\to 0$ at the end is ambiguous. Yes, you and I know that $n$ is tending to infinity. To an elementary student, however, this is notation soup, and rather confusing. This could be very simply improved by writing $$ \left( \frac{\mathrm{e}}{3} \right)^n \stackrel{n\to \infty}{\longrightarrow} 0,$$ or by using the notation familiar to elementary students, i.e. $$ \lim_{n\to\infty}\left( \frac{\mathrm{e}}{3} \right)^n = 0.$$

Again, I think that the overall problem (and the reason for my downvote of the original answer) is that the solution is not written at a level that is useful for anyone who might actually ask the original question. I would hazard that most of the folk who follow your answer already knew how to deal with the question, and that most of the people who couldn't answer the question are likely not helped by your answer.

One final note: you claim that

Despite all my efforts to explain these basic concepts, the answer has been finally deleted by moderators without any good explanation.

None of the explanation that you sought to provide was in the answer. The answer did not stand alone as an answer, but required a careful reading of the attached comments. If an answer is not clear and requires explanation, then (it seems to me) the correct thing to do is edit the clarification into the answer, and not rely upon the (more ephemeral) comments to get the job done.

Answer 2

Since I down-voted and voted to delete your answer I might at least say why I did that.

First of all, you are skipping probably the hardest part of question, that $$n < \left(\frac e2\right)^n$$

when $n$ is large. And you are (i) hiding this in the first inequality (instead of isolating it and pointing it clearly) and (ii) choose the constant $e$ in a confusing way: all you need is something less than $3$, and adding this $e$ here makes people harder to understand what is going on.

Secondly, you are using the term "definitively". I hope you realize now that this is not a very common term.

I don't know why you write this answer. The OP seems to be a newbie in this stuff, if someone wants to help, I hope they at least make their answer clear enough. I think all the others answers are okay in doing that.

Because of that, I found your answer the least useful of all. Then I see your comment, when Myridium asks for clarification:

"If one person downvote for this trivial point, I presume he is not much skilled in limits calculation."

I then had the feeling that you are not likely to edit your answer and thus I voted to delete.