TITLE: Evaluating $\lim_{n\rightarrow\infty} (\frac{(1+\frac{1}{n})^n}{e})^n$
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: 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 $\frac{3713}{3600 \sqrt{e}}=0.625569$.