Supremum of $(1-1/n^2)^n$

Question

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.

Answer 1

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 $1$.

Answer 2

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

Answer 3

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.

Answer 4

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 $99.5042$