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

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.

• The answer was deleted by high reputation users(3 of them) not the diamond moderators. – clark Jan 7 '18 at 8:47
• BTW . 2-finger typists like me prefer n \to \infty over n \rightarrow \infty. – DanielWainfleet Jan 7 '18 at 10:24
• I answered that question too - or at least gave a method to obtain an answer using the binomial theorem. I think people had more problem with the question than any of the answers, and wanted the question improved before anyone gave a decent answer. – Mark Bennet Jan 7 '18 at 11:54
• @gimusi Here are my 2 cents about the answer: It's not a good answer as it stands. It's not going to be at all useful to an asker of this level, because someone who doesn't know that $n \cdot (2/3)^n \to 0$ is going to have no idea how to prove that $n < (e/2)^n$, or why you made that estimate. The length of the answers here, as well as the length of the comment chain on the original answer (before it went off the rails...) support this. Whether it merits deletion... I'm not so sure. But it probably does, because it's not a good answer to a poorly posed question. – user296602 Jan 7 '18 at 18:55
• @user296602 Thanks for your comment. Maybe you are right but my intention was to receive a feedback by who posed he OP in order to better explain the answer and how to deal with this kind of limits. – user Jan 7 '18 at 19:00
• Is there some reason you linked to Zaharyas' answer (and not to your answer or to the question)? – Martin Sleziak Jan 7 '18 at 19:01
• @gimusi AmWhy is not a moderator of the site, and it's simply an informative comment (because users with less than 10K reputation will not be able to see your answer). And if your intention is to get feedback from the asker, then it's appropriate to do that in the comments. An answer should (generally) not be an invitation to dialogue, but a self-sufficient answer. – user296602 Jan 7 '18 at 19:02
• Possible duplicate of Requests for Reopen & Undeletion Votes, etc. (volume 01/2015 - ) [current version] – Namaste Jan 7 '18 at 19:10
• You have repeatedly been given suggestions, non of which you acknowledge as worthwhile. It seems you have done nothing more than solicit undeletion votes, for which two users thus far have granted you. You want to seriously improve your answer? Then do some rereading, and seriously read the answers below, and importantly, make sure you speak in terms/language that the asker can understand, this time. Else, upon its undeletion, it will again face deletion unless you do, indeed, improve your answer. – Namaste Jan 7 '18 at 19:31
• @amWhy The general reopen request thread is supposed to avoid any controversy, and there's clearly substantial disagreement about the fate of this post. So I don't think closure as duplicate is appropriate here. – user296602 Jan 7 '18 at 19:48
• Changing "definitively" to "eventually" will make the two answers below meaningless. And "definitively" is the word you used in the original post, there is no point to change it here. – user99914 Jan 7 '18 at 20:59
• Note that comment are not permanent, they can be deleted by moderators or the community (indeed, the comment about the definition of "definitively" is gone now, as you can see in your deleted answer). Everything that is essential should be edited to the answer. @gimusi – user99914 Jan 7 '18 at 21:08
• gimusi: For your information: I deleted both Daniel Wainfleet's and Siong Thye Goh's answers. Because IMO they were answers to a question in main - not answers to a question in meta. – Jyrki Lahtonen Jan 8 '18 at 7:59
• @Myridium It is not the same thing in the spirit of my proof, indeed if I plug in this inequality I obtain that $n2^n/3^n<1$ which is not conlusive for the application of squeez theorem. – user Jan 9 '18 at 10:30
• Downvoting is one thing and deletion is another. It's shameful that OP should have to come here and make this reasonable defense and be treated this way, if perfection is the standard why not delete 95% of the site? – Dan Brumleve Jan 11 '18 at 21:50

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.

• This is a great explanation of the issues with the post, and the last paragraph is a very succinct argument for why this should be deleted. Only thing I'd suggest adding (for the benefit of the asker, perhaps) is that there were no moderators involved in the deletion vote - all four of the users who cast deletion votes are normal users, not diamond moderators. – user296602 Jan 7 '18 at 19:10
• Thanks for your detailed explanation, I understand your point of view and it can of course justify a downvote, but I really do not understand the motivation for its deletion and also the others deletion here of the answers from other expert users as math.meta.stackexchange.com/users/306553/siong-thye-goh with a different point of view. – user Jan 7 '18 at 19:10
• @gimusi this answer seems to have been given on main and treat the question like a mathematical question. This is not really an answer to a meta question. It is not clear they have a different point of view (though they might). They explained what is missing in your answer; you should note that they do find quite a bit missing in your answer. – quid Jan 7 '18 at 19:29
• @quid Initially the OP was posted on MSE and after it has been migrated here not for my choice. Indeed they had differnt point of view in the sense that they explained what something was missing but they considered the original answer an acceptable answer, notably Siong Thye Goh wrote: "I think the remaining argument makes sense besides the gap". – user Jan 7 '18 at 19:37
• @gimusi this is not saying your answer "as is" is acceptable. It could also mean that if you close the gap, then it would be acceptable, otherwise it might not. I am pretty sure though that yes you will find many user that think your answer should not be deleted, because many users think that no answer at all should be deleted (except for spam and alike). – quid Jan 7 '18 at 19:42
• gimusi: you say "Thanks for your detailed explanation, I understand your point of view and it can of course justify a downvote, but I really do not understand the motivation for its deletion." You asked how you can improve your post. That is not a question about "other's motivation for deleting your post," although it has been provided. Now, are you looking for ways to improve your deleted post, or are you just downright mad as heck, holding firm to some belief of your supposed "brilliance" and will forever be unhappy unless we all agree that you are "right" and we're all wrong?. – Namaste Jan 7 '18 at 20:00
• If so, then you are not, sincerely, asking for ways to improve your answer, and hence it belongs, if anywhere, on the thread I linked to below your posted rant? – Namaste Jan 7 '18 at 20:01

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.

• Thanks for your comment. I really think that anyone here must be free to express his point of view with comments, upvoting and downvoting but to eliminate an answer just because it does not satisfy the preferences and tastes of some people seems unreasonable to me. Instead, it should be more useful and appropriate to comment on the answer to try to reason together and maybe improve it and eventually learn something more. – user Jan 7 '18 at 19:19
• @gimusi did you read the last sentence of the answer? It seems it addresses what you write. – quid Jan 7 '18 at 19:24
• @quid John Meta didn't make any comment to my original answer, how could I presume his thought about the prossible improvements if he downvote without explain. Anyway I have to repeat my question here is not about downvotes but I think that deleting the aswer we don't like is not a correct way to teach simething here. – user Jan 7 '18 at 19:42
• Why, @gimusi, do you dismiss out of hand every single critique or suggestion others have given, with which you may be able to improve your answer? You claim you are not trying to solicit undelete votes, nor solicit upvotes, just trying to learn how to improve your answer. But you reject, dismiss, or even belittle helpful suggestions from users who are trying to help you do this. At this point, I can only conclude that you remain convinced that if only everyone "knew what you know", there'd be no problem. But, it seems clear to me that the problem is yours and your unwillingness to listen. – Namaste Jan 7 '18 at 19:49
• @gimusi The hole in your answer was pointed out less than five minutes after you posted it. What would be the point in repeating that comment? – Daniel Fischer Jan 7 '18 at 20:20
• @DanielFischer I've received only some feedback from Myridium and I've replied to his request of clarifications. – user Jan 7 '18 at 20:26
• @gimusi And you've made pretty clear that you had no intention of plugging the hole. That's the point. – Daniel Fischer Jan 7 '18 at 20:28
• @DanielFischer I claimed that it was not the main point in my proof and that it is and independent basic result. Indeed I think that we don't need to prove everytime all the math facts to show that a single result is true. Certainly I would have given more explanations to RAVE if he/she had asked to me some more detail on the proof. No one questioned the correctness of this assumption with the exception of Myridium who wasn't aware about the mening of the term "definitively" . – user Jan 7 '18 at 20:38
• @gimusi Anybody for whom that result is basic wouldn't have any problems proving that $\frac{n2^n}{3^n} \to 0$. What results one can reasonably use without proof very much depends on context. In the given context, using that an increasing geometric sequence grows faster than the index without proof is not reasonable, since the end goal immediately follows from that. And "definitively" is not a standard expression. I suppose it's the literal translation of the word used in your native language, in English the usual word for "from some $n_0$ on" is "eventually". – Daniel Fischer Jan 7 '18 at 20:46
• @DanielFischer Yes it is indeed! Thanks for the next limit of this kind I will use the term "eventually". I understand your point of view and appreciate very much your kindly and inteligent manner to explain it. Anyway I think that delete the answer is not a good way to exchange knoledge here. Instead such comments from person skilled like you are very useful. – user Jan 7 '18 at 20:52
• @gimusi: asking for improvement is good and should be encouraged. But you should leave your question in meta as what it was so that you would not invalid people's helpful comments/answers here. – Jack Jan 8 '18 at 0:38
• @Jack Yes I agree, I've finally added a final note in order to be clear that is an EDIT made after the discussion here. Thanks. – user Jan 9 '18 at 9:14
• @gimusi - I would like to add that I did tell you my understanding of the definition of the word "definitively" which, at the time, did not prompt you to change it. – Myridium Jan 9 '18 at 10:10