Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$

Question

I want to show that $$ \lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right) = \lim_{n \to \infty} \prod_{k=0}^{n-1} \left(1-\frac{n}{n^2 - k} \right) = \mathrm{e}^{-1} $$

I know that $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ and so my attempt was to write

$$ \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)\\ = \underbrace{\left(1-\frac{1}{n} \right)^n}_{\to \text{e}^{-1}} \cdot \underbrace{\frac{1-\frac{n}{n^2-1}}{1-\frac{1}{n}}}_{\rightarrow 1} \cdot \ldots \cdot \underbrace{\frac{1-\frac{n}{n^2-n+1}}{1-\frac{1}{n}}}_{\rightarrow 1} $$ which I thought would solve my problem. But after a second look I see that splitting the limits in the product cannot be allowed. This would be the same nonsense as $$ \underbrace{\left(1-\frac{n}{n^2} \right)}_{\rightarrow 1} \underbrace{\left(1-\frac{n}{n^2-1} \right)}_{\rightarrow 1} \cdot \ldots \cdot \underbrace{\left(1-\frac{n}{n^2-n+1} \right)}_{\rightarrow 1}.$$

Maybe anybody can make the situation clear to me.

Answer 1

We have, by a simple comparison, $ \left( 1-\frac{n}{n^2-n} \right)^n \le \prod_{i=0}^{n-1} \left( 1-\frac{n}{n^2-i} \right) \le \left( 1-\frac{1}{n} \right)^n $. The limit of the left and right hand products as $ n \rightarrow \infty $ is $ \frac{1}{e} $, so by the Squeeze theorem, the limit of the middle product is also $ \frac{1}{e} $ as you wanted.

Answer 2

$$ \left(1-\frac{n}{n^2 - n + 1} \right)^n \leq \prod_{k=0}^{n-1} \left(1-\frac{n}{n^2 - k} \right) \leq \left(1-\frac{n}{n^2} \right)^n $$ $$\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right)^n = \frac{1}{e}$$

$$\begin{align*} \lim_{n \to \infty} \left(1-\frac{n}{n^2 - n + 1} \right)^n &= \lim_{n \to \infty} \left(1-\frac{1}{n - 1 + 1/n} \right)^n \\ &= \lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \frac{1}{e} \end{align*}$$ so $\lim_{n \to \infty} \prod_{k=0}^{n-1} \left(1-\frac{n}{n^2 - k} \right) = \frac{1}{e}$.

Answer 3

May be another way to see the limit and how it is approached.

Let $$u_i=1-\frac 1{n^2-i}\implies P_n=\prod_{i=0}^{n-1} u_i$$ Take logarithms $$\log(P_n)=\sum_{i=0}^{n-1} \log(u_i)$$ Now, use Taylor series for large values of $n$ $$\log(u_i)=-\frac{1}{n}-\frac{1}{2 n^2}-\frac{i+\frac{1}{3}}{n^3}-\frac{i+\frac{1}{4}}{n^4}+O\left(\frac{1}{n^5 }\right)$$ Now, compute the sums $$\log(P_n)=-1-\frac{1}{2 n}+\frac{1-3 n}{6 n^2}+\frac{1-2 n}{4 n^3}+\cdots=-1-\frac 1 n-\frac 1 {3n^2}+\frac 1 {4n^3}+\cdots$$ Going back to the exponential $$P_n=\frac 1e\Big(1-\frac{1}{n}+\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right) \Big)$$