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We know that $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ and so I thought the approach to evaluating the limit in the question would be to just use this fact and substitute it into the numerator. This approach would tell us the above limit evaluates to $1$. However, that does not seem to be the correct limiting value. In fact, it evaluates to $\frac {1}{\sqrt{e}}$. Why is this so?

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    $\begingroup$ Why the close vote? This is a well-posed question that shows some work/ideas. $\endgroup$ Commented May 14, 2020 at 0:22
  • $\begingroup$ The answer is already there in your first line. $ 1 +\frac 1 n \to 1$ but $(1+\frac 1 n)^{n}$ does not tend to $1$. (I did not downvote). $\endgroup$ Commented May 14, 2020 at 0:28
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    $\begingroup$ @KaviRamaMurthy, the argument the OP is making is that the expression inside the outer parentheses should tend to 1: $\frac{1}{e} \lim_{n \to \infty} (1+\frac{1}{n})^n = 1$. Nowhere does he/she refer to $\lim_{n \to \infty}(1 + \frac{1}{n}) = 1$. The question is essentially why you can't take the limit on the inside of the parentheses and then get $1^n \to 1$. The OP is aware that this argument is wrong, but doesn't know why. $\endgroup$
    – sasquires
    Commented May 14, 2020 at 0:34

7 Answers 7

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This is a companion to Eevee Trainer's answer: The same (il)logic that says

$$\left((1+1/n)^n\over e\right)^n\to\left(e\over e\right)^n=1^n=1$$

would also say

$$\left(1+{1\over n}\right)^n\to(1+0)^n=1^n=1$$

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    $\begingroup$ You make an excellent point! $\endgroup$ Commented May 14, 2020 at 3:26
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Let $x=\frac1n$. Using $$ \ln(1+x)=x-\frac12x^2+O(x^3)$$ one has \begin{eqnarray} &&\lim_{n\rightarrow\infty} \ln\bigg(\frac{(1+\frac{1}{n})^n}{e}\bigg)^n\\ &=&\lim_{n\rightarrow\infty} n\bigg(n\ln(1+\frac{1}{n})-1\bigg)\\ &=&\lim_{n\rightarrow\infty} \frac{n\ln(1+\frac{1}{n})-1}{\frac1n}\\ &=&\lim_{x\rightarrow0} \frac{\frac1x\ln(1+x)-1}{x}\\ &=&\lim_{x\rightarrow0} \frac{-\frac12x+O(x^2)}{x}\\ &=&-\frac12. \end{eqnarray} So $$ \lim_{n\rightarrow\infty} \bigg(\frac{(1+\frac{1}{n})^n}{e}\bigg)^n=e^{-1/2}. $$

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Since everyone else has decided to cover ways to calculate the limit (despite a comment from you mentioning that you want to know why you're wrong and not how to solve the exercise), I'll answer focusing on that. First, simplifying, you have

$$\lim_{n \to \infty} \frac{ ((1+1/n)^n)^n }{e^n}$$

In replacing the top expression with $e^n$, you implicitly assume that you can take the limit inside as so, with your substitution in blue:

$$\lim_{n \to \infty} \left( \left( 1 + \frac 1 n \right)^n \right)^n = \left(\color{blue}{\lim_{n \to \infty} \left( 1 + \frac 1 n \right)^n} \right)^n =\color{blue}{e}^n$$

However, you have a dependence on $n$ on the outer parentheses, and thus this step is not justified. You can only move a limit inside a (continuous) function when you're not suddenly moving a dependence on $n$ to the outside.

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Just for your curiosity.

We can have much more than the limit itself composing Taylor series $$a_n=\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right)^n\implies \log(a_n)=n^2 \log\left(1+\frac{1}{n}\right)-n$$Now, using Taylor expansion $$\log(a_n)=n^2\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}-\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right)\right)-n$$ $$\log(a_n)=-\frac{1}{2}+\frac{1}{3 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$ $$a_n=e^{\log(a_n)}=\frac 1 {\sqrt e}\left(1+\frac{1}{3 n}-\frac{7}{36 n^2}\right)+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

Moreover, this gives you a shortcut method for a quick evaluation of $a_n$ even for small values of $n$. For example, the exact calculation of $a_{10}= 0.625639$ while the above truncated expansion gives $\frac{3713}{3600 \sqrt{e}}=0.625569$.

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Since $ 1^{\infty} $ is an indeterminate form, we can not just say that "since $ \lim\limits_{n\to +\infty}{\left(1+\frac{1}{n}\right)^{n}}=\mathrm{e} $, the limit of the ratio raised to the $ n^{\mathrm{th}} $ power will be just $ 1 $".

Let's proceed without using either L'Hospital's rule or series expansion :

Let $ n $ be a positive integer, we have :

\begin{aligned}\lim_{n\to +\infty}{\left(\frac{\left(1+\frac{1}{n}\right)^{n}}{\mathrm{e}}\right)^{n}}&=\lim_{n\to +\infty}{\mathrm{e}^{-n}\left(1+\frac{1}{n}\right)^{n^{2}}}\\ &=\lim_{n\to +\infty}{\exp{\left(-n+n^{2}\ln{\left(1+\frac{1}{n}\right)}\right)}}\\ &=\lim_{n\to +\infty}{\exp{\left(-\frac{\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}}{\frac{1}{n^{2}}}\right)}}\end{aligned}

Now let $ x\geq 0 $, note that : \begin{aligned} \frac{x-\ln{\left(1+x\right)}}{x^{2}}&=\int_{0}^{1}{\frac{1-y}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\\ &=\int_{0}^{1}{\left(1-y\right)\mathrm{d}y}-\int_{0}^{1}{\left(1-y\right)\left(1-\frac{1}{\left(1+xy\right)^{2}}\right)\mathrm{d}y}\\ &=\frac{1}{2}-x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y} \end{aligned}

Since $ x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\leq x\left(2+x\right)\int_{0}^{1}{y\,\mathrm{d}y}=\frac{x\left(2+x\right)}{2}\underset{x\to 0}{\longrightarrow}0 $, we get that : $$ \lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $$

Hence : $$ \lim_{n\to +\infty}{\frac{\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}}{\frac{1}{n^{2}}}}=\frac{1}{2} $$

Which means : $$ \lim_{n\to +\infty}{\left(\frac{\left(1+\frac{1}{n}\right)^{n}}{\mathrm{e}}\right)^{n}}=\lim_{n\to +\infty}{\exp{\left(-\frac{\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}}{\frac{1}{n^{2}}}\right)}}=\exp{\left(-\frac{1}{2}\right)} $$

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Rewrite the function as $\phi(n) = e^{\log \phi(n)} = e^{n^2 \log (1+\frac{1}{n}) - n} = e^{n^2(\frac{1}{n} -\frac{1}{2n^2} + O(\frac{1}{n^3}))- n} = e^{-\frac{1}{2} + O(\frac{1}{n})} = e^{-\frac{1}{2}}$

I used Maclaurin series expansion here

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  • $\begingroup$ You forgot the 1/2 on the second term in the Maclaurin series (which is what gives the answer that the reader found numerically). You also forgot to multiply the $n^2$ through on the other terms. $\endgroup$
    – sasquires
    Commented May 14, 2020 at 0:41
  • $\begingroup$ @sasquires: fixed $\endgroup$
    – Alex
    Commented May 14, 2020 at 0:52
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$1^\infty$ is an indeterminate form. Taking logarithm, we get

$$n\Bigl(\ln((1+\frac 1n)^n)-1\Bigr)=$$

$$n\Bigl(n\ln(1+\frac 1n)-1\Bigr)=$$

$$n\Bigl(n(\frac 1n-\frac{1}{2n^2}+\frac{1}{n^2}\epsilon(n))-1\Bigr)=$$

$$\frac{-1}{2}+\epsilon(n)$$

the limit is $$\sqrt{\frac 1e}$$

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