Supremum of $(1-1/n^2)^n$ I want to find the supremum of $$(1-1/n^2)^n$$ where $n\in \mathbb{N}$ and prove it.
Since $(1-1/n^2)<1$ raising this to the power will still be less than one. So we showed that $(1-1/n^2)^n<1$. But how can I show that this is the smallest upper bound? The definition of the supremum says that any smaller number than the supremum cannot be an upper bound. So if I take a number $1-\epsilon$ $\epsilon>0$ then there must exist an $n$ such that $$1-\epsilon<(1-1/n^2)^n$$ So we have to find that $n$, but solving this inequality seems highly nontrivial and I am not sure how to proceed from here.
 A: Basic approach. First, show that $(1-a)(1-b) > 1-a-b$ for $a, b > 0$.
Second, use this plus induction to show that $\left(1-\frac{1}{n^2}\right)^n > 1-\frac1n$ for positive integer $n$.
Finally, combine this with the fact that $\left(1-\frac{1}{n^2}\right)^n < 1$ for positive integer $n$ to show that the supremum must be $1$.
A: Let $a_n=\left(1-\frac1{n^2}\right)^n$.  Note from Bernoulli's Theorem that
$$1-\frac1n\le a_n\le 1$$
Hence, by the squeeze theorem $\lim_{n\to\infty}a_n = 1$.  And this is sufficient to establsih that $\sup_n a_n=1$.


BONUS MATERIAL:
Interestingly, we can also show that $a_n$ is increasing monotonically.  To do so, we form the ratio $\frac{a_{n+1}}{a_n}$ and show that this is always greater than $1$.  Proceeding, we have
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{\left(1-\frac1{(n+1)^2}\right)^{n+1}}{\left(1-\frac1{n^2}\right)^n}\\\\
&=\frac{\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}}{\left(\frac{(n+1)(n-1)}{n^2}\right)^n}\\\\
&=\frac{(n+1)(n-1)}{n^2}\left(1+\frac{2n+1}{(n-1)(n+1)^3}\right)^{n+1}\\\\
&\ge \frac{(n+1)(n-1)}{n^2}\left(1+\frac{2n+1}{(n-1)(n+1)^2}\right)\\\\
&=\frac{n^2+n+1}{n(n+1)}\\\\
&\ge 1
\end{align}$$
So, $a_n$ is monotonically increasing and bounded above by $1$ and is convergent.
A: Here is another approach that I find fun and quite basic.
Let us write $(a_n)_{n \ge 1}$ where
$$
a_n := \left(1-\frac{1}{n^2}\right)^n.
$$
You noted that $a_n \le 1$ for each $n \ge 1$. So, it suffices to show that $a_n \to 1$ as $n \to \infty$. To prove this, let us define the sequence $(b_n)_{n \ge 1}$ where
$$
b_n := 1-\frac{1}{n^2},
$$
meaning $a_n = b_n^n$ for each $n \ge 1$. For $n \ge 2$, we observe by the triangle inequality that
$$
|a_n-1| = |b_n^n-1| \le |b^n-b_n|+|b_n-1| = |b_n||b_n^{n-1}-1|+|b_n-1| \le |b_n^{n-1}-1|+|b_n-1|
$$
since $|b_n| < 1$. Continuing on if $n \ge 3$, we may apply the same trick to obtain
$$
|a_n-1| \le |b_n^{n-1}-1|+|b_n-1| \le |b^{n-1}-b_n|+2|b_n-1| \le |b_n^{n-2}-1|+2|b_n-1|.
$$
Formalizing this reasoning via induction (try this yourself), one can show that
$$
|a_n-1| \le n|b_n-1| = |nb_n-n|
$$
for all $n \ge 1$. So, we have reduced the problem to showing that $nb_n-n \to 0$ as $n \to \infty$. Yet, we compute
$$
nb_n-n = n\left(1-\frac{1}{n^2}\right)-n = -\frac{1}{n}
$$
which clearly goes to $0$ as $n \to \infty$. So we are done!
edit: fixed typo
A: Solving for $n$ the inequality $$1-\epsilon<\left(1-\frac{1}{n^2}\right)^n$$ is not so bad. Take logarithms
$$\log(1-\epsilon)<n \log\left(1-\frac{1}{n^2}\right)=-\frac{1}{n}+O\left(\frac{1}{n^3}\right)$$
$$n > -\frac 1 {\log(1-\epsilon)}$$ Trying with $\epsilon=0.01$ gives, as a real, $n>99.4992$ while the exact solution is $99.5042$
