# How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$?

I was reading about how completeness is required for limits. And I came across this:

the sequence $a_n=(1+1/n)^n$ is monotonically increasing and bounded by 3 and so we expect it to converge, but that it does not converge within $\mathbb{Q}$. More generally it stands to reason that any sequence of real numbers which is increasing and bounded must converge to some real number. This is a consequence of completeness of $IR$

My question is: How is the mentioned sequence monotonically increasing and bounded by $3$ ?

• you can prove it by induction Commented Oct 5, 2013 at 6:41
• You cannot generally prove convergence statements using induction. Induction only shows things to be true for finite n. Commented Oct 5, 2013 at 7:02
• @Jebruho: If $(a_n)$ is a sequence, the statements $a_{n+1}\geq a_n$ and $a_n\leq 3$ are statements about integers, hence may in principle be proved by induction. For convergence in $\mathbb R$ one then needs to invoke completeness, but GTX OC is not asking about that part. Commented Oct 5, 2013 at 7:37
• Commented Oct 5, 2013 at 7:38

The monotonicity follows from the AM-GM inequality for $n+1$ points. Taking $x_1 = x$ and $x_2 = \cdots = x_{n+1} = y$, we get $$\sqrt[n+1]{xy^n} \leq \frac{x+ny}{n+1}.$$ In particular, taking $x = 1$ and $y = 1+\frac{1}{n}$ yields $a_n \leq a_{n+1}$ ($n \geq 1$).

Now as another special case, take $x = 1$ and $y = 1-\frac{1}{n}$ to get $$b_n \geq b_{n+1} \qquad (n \geq 1),$$ where $b_n := \left( 1-\frac{1}{n}\right)^{-n}$. We let you verify that $$b_n = a_{n-1} \left( 1+\frac{1}{n-1}\right) \geq a_{n-1}, \qquad (n \geq 2).$$ Hence $$a_1 \leq a_2 \leq a_3 \leq \cdots \leq b_3 \leq b_2 \leq b_1$$ and we deduce that $a_n \leq b_m$ for all $n, m \in \mathbb{N}$. In particular, taking $m = 6$, we get $$a_n \leq \left( \frac{6}{5} \right)^6 \leq 3.$$

• Can't we show that it is increasing monotonically by differentiating $a_n$ with respect to n and checking whether it is always positive for all n $\leq 1$ ? Commented Oct 5, 2013 at 7:49
• Well you can't differentiate with respect to $n$ because $n \in \mathbb{N}$. However you could try and differentiate the function $(1+1/x)^x$ on $x > 0$. Then if you can show it is always $\geq 0$ on $x > 0$ that would do it. Commented Oct 5, 2013 at 7:57

From Rudin's PMA Theorem $$3.31$$, by the Binomial theorem,

$$t_n=\left( 1 + \frac{1}{n} \right)^n$$

$$= 1 + 1 + \underbrace{\frac{1}{2!}\left(1-\frac{1}{n}\right)}_{\text{term } 2} + \underbrace{\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}_{\text{term } 3} + \ldots + \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right).$$

Since each bracket increases as $$n$$ increases, each term increases as $$n$$ increases also, because products and sums of increasing positive functions is also increasing.

Therefore, for all $$n\geq 2,$$ and all $$2\leq k \leq n,$$ the $$k$$-th term of $$t_{n+1}$$ is greater than the $$k$$-th term of $$t_n.$$

So we see that $$\left( 1 + \frac{1}{n} \right)^n$$ is increasing.

Furthermore, the Binomial expansion above is

$$\leq 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} = 1 + 1 + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \frac{1}{1\cdot 2\cdots n}$$

$$< 1+1+ \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}} <3.$$