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$\lim \limits_{n \to \infty\ } \Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n$

I would like to replace $(1+\frac{1}{n})^n$ by $e$, and then $\frac{2,7}{e}<1$, so $\lim \limits_{n \to \infty\ } \Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n=0$, and I will get correct result, but I think this replacement is inadmissible.

I'm looking for the easiest way ( without advanced tools)

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  • $\begingroup$ The replacement is inadmissible but the idea can be used in a rigorous manner thanks to squeeze theorem. See my answer for details. $\endgroup$
    – Paramanand Singh
    Nov 16 '18 at 4:01
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Yes the limit is zero indeed by root test

$$\sqrt[n]{\Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n}= \frac{2,7}{(1+\frac{1}{n})^n}\to\frac{2.7}e<1$$

As an alternative, to make your way rigorous we need to observe that eventually

$$\frac{2,7}{\left(1+\frac{1}{n}\right)^n}\le \frac{2,7}{\frac{e+2.7}{2}}<1$$

and conclude by squeeze theorem

$$\left(\frac{2,7}{\left(1+\frac{1}{n}\right)^n}\right)^n\le \left(\frac{2,7}{\frac{e+2.7}{2}}\right)^n\to 0$$

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  • $\begingroup$ I can't use methods which haven't been proven at my lectures. $\endgroup$ Nov 15 '18 at 22:51
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    $\begingroup$ It's probably also worth noting that the proof in the question doesn't work, for precisely the same reason that $\left(1+\frac{1}{n}\right)^n \to \left(1 + 0\right)^n \to 1$ doesn't. $\endgroup$ Nov 15 '18 at 22:52
  • $\begingroup$ @matematiccc I've added a secon way to prove that. $\endgroup$
    – user
    Nov 15 '18 at 22:53
  • $\begingroup$ Could you explain why $\frac{2,7}{\left(1+\frac{1}{n}\right)^n}\le \frac{2,7}{\frac{e+2.7}{2}}$? I mean, why we can do it. $\endgroup$ Nov 15 '18 at 23:05
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    $\begingroup$ @matematiccc Since $\left(1+\frac{1}{n}\right)^n \to e$ eventually we have that it is greater that $2.7$ but smaller that $e$ (since the sequence is increasing). $\endgroup$
    – user
    Nov 15 '18 at 23:09
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It sounds like you already know that $(1 +1/n)^n \to e$. By the definition of a limit: for all $\epsilon > 0$, there exists $N > 0$ so that $|(1 + 1/n)^n - e| < \epsilon$ for all $n \geq N$.

Consider $\epsilon = 1/100$. Then, $|(1 + 1/n)^n - e| < 1/100$ for $n \geq N$ for some $N$. Thus, $(1 + 1/n)^n \geq 2.708$ for $n \geq N$.

Finally we have $2.7 / (1 +1/n)^n \leq 2.7/2.708$, and hence $\left( 2.7 / (1 +1/n)^n\right)^n \leq \left(2.7/2.708\right)^n$ for $n \geq N$.

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  • $\begingroup$ This is same my answer which was posted a few minutes earlier. $\endgroup$
    – Paramanand Singh
    Nov 16 '18 at 4:02
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    $\begingroup$ @ParamanandSingh Yes, it was posted 40 second earlier. If you are interesting in nit-picking, then user gimusi posted this idea as a comment 5 hours ago. $\endgroup$
    – JavaMan
    Nov 16 '18 at 4:10
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You need two results here. First is that the limit of $(1+(1/n))^n$ exists and is commonly denoted by $e$. Second is the fact that $e>2.7$.

Next note that if $a_n=2.7/(1+(1/n))^n$ then by above two facts the sequence $a_n\to l$ where $0<l<1$. Let's choose a specific number $\epsilon>0$ such that $$0<l-\epsilon<l+\epsilon<1$$ This we can choose $\epsilon$ to be any positive number less than $\min(l, 1-l)$.

Since $a_n\to l$ it follows by definition of limit that there is a positive integer $m$ such that $$l-\epsilon <a_n<l+\epsilon, \forall n\geq m$$ Note that each term of the inequality is positive and hence if we raise each term to $n$'th power we get $$(l-\epsilon) ^n<a_n^n<(l+\epsilon) ^n, \forall n\geq m$$ Applying squeeze theorem we get the desired limit as $0$ because both $l-\epsilon, l+\epsilon $ lie between $0$ and $1$.

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You're right – one cannot replace only a part of an expression with its limit.

The simplest way consists in determining the limit of the log, using Taylor's formula at order $12$: \begin{align} n\log(2.7)-n^2\log\Bigl(1+\frac1n\Bigr)&=n\log(2.7)-n^2\biggl(\frac1n-\frac1{2n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggr) \\ &=n\log(2.7)-n+\frac12+o(1)\\ &=n(\underbrace{\log 2.7-1}_{<\,0})+\frac12+o(1)\bigg]\to -\infty. \end{align}

Without Taylor's formula:

From the inequalities $\;1-x<\dfrac1{1+x}<1$ $(x>0)$, you can deduce with the mean value theorem that $$x-\frac{x^2}2<\log(1+x)<x\quad\forall x>0$$ so that \begin{align} n\log(2.7)-n^2\log\Bigl(1+\frac1n\Bigr)&<n\log(2.7)-n^2\biggl(\frac1n-\frac1{2n^2}\biggr) = n(\underbrace{\log 2.7-1}_{<\,0})+\frac12 \end{align} The same conclusion as above follows by the comparison theorem.

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  • $\begingroup$ Thanks, I understand it, but I can use at most squeeze theorem. $\endgroup$ Nov 15 '18 at 23:04
  • $\begingroup$ I've added a proof by the comparison theorem (no squeeze theorem required here since the limit of the log is $-\infty$). $\endgroup$
    – Bernard
    Nov 15 '18 at 23:29

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