# Evaluate $\lim \limits_{n \to \infty\ } \Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n=0?$

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

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)

• The replacement is inadmissible but the idea can be used in a rigorous manner thanks to squeeze theorem. See my answer for details. Nov 16 '18 at 4:01

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

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

• I can't use methods which haven't been proven at my lectures. Nov 15 '18 at 22:51
• 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. Nov 15 '18 at 22:52
• @matematiccc I've added a secon way to prove that.
– user
Nov 15 '18 at 22:53
• 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. Nov 15 '18 at 23:05
• @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).
– user
Nov 15 '18 at 23:09

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 $$n \geq N$$.

• This is same my answer which was posted a few minutes earlier. Nov 16 '18 at 4:02
• @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. Nov 16 '18 at 4:10

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 $$0. Let's choose a specific number $$\epsilon>0$$ such that $$0 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 $$l-\epsilon Note that each term of the inequality is positive and hence if we raise each term to $$n$$'th power we get $$(l-\epsilon) ^n Applying squeeze theorem we get the desired limit as $$0$$ because both $$l-\epsilon, l+\epsilon$$ lie between $$0$$ and $$1$$.

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 $$x-\frac{x^2}2<\log(1+x)0$$ so that \begin{align} n\log(2.7)-n^2\log\Bigl(1+\frac1n\Bigr)& The same conclusion as above follows by the comparison theorem.

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