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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 $3$ ?

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  • $\begingroup$ you can prove it by induction $\endgroup$
    – hchengz
    Commented Oct 5, 2013 at 6:41
  • $\begingroup$ You cannot generally prove convergence statements using induction. Induction only shows things to be true for finite n. $\endgroup$
    – Jebruho
    Commented Oct 5, 2013 at 7:02
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    $\begingroup$ @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. $\endgroup$ Commented Oct 5, 2013 at 7:37
  • $\begingroup$ Related: math.stackexchange.com/questions/167843/…, math.stackexchange.com/q/220723 $\endgroup$ Commented Oct 5, 2013 at 7:38

2 Answers 2

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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 $$ a_n \leq \left( \frac{6}{5} \right)^6 \leq 3. $$

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  • $\begingroup$ 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$ ? $\endgroup$
    – GTX OC
    Commented Oct 5, 2013 at 7:49
  • $\begingroup$ 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. $\endgroup$
    – Amateur
    Commented Oct 5, 2013 at 7:57
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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

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

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

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