Simple proof that $(1 + 1/n)^n$, $n \in \mathbb N$, is bounded above?

Question

Here is the context of the problem. I'd like to prove that the sequence $$a_n = \left( 1 + \frac 1n \right)^n$$ converges using the monotone convergence theorem. It is straightforward using Bernoulli's inequality to show that $\{a_n\}$ is increasing. To conclude the sequence converges it suffices to show it is bounded above.

The toolbox available is somewhat limited. Just the basic ordered field properties, Bernoulli's inequality, etc. No exponential functions or logarithms are available.

An argument that is essentially identical to the proof that $\{a_n\}$ is increasing shows that the sequence $$ b_n = \left( 1 + \frac 1n \right)^{n+1}$$ is decreasing. Since $\{b_n\}$ is trivially bounded below it converges, and an application of the limit law shows $\{a_n\}$ converges to the same limit. This means I have a simple proof that $\{ a_n\}$ converges. I'm curious if I am missing an easy way to work only with $\{a_n\}$. My question is that of the title:

Is there a simple proof that $\{a_n\}$ is bounded above?

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Edited to add: thanks for the suggestions so far. The precise context of the problem is as an example in a real analysis class. The students are aware of the fact that the line is a complete ordered field, have seen the definition of a convergent sequence, have proved some limit laws, and have just seen the proof of the monotone convergence theorem. The binomial theorem and anything involving series aren't yet known. I could work through the proof of the binomial theorem, but I'm looking to see if there happens to be a really simple proof of boundedness of $\{a_n\}$, not much more difficult than the proof that $\{b_n\}$ is bounded below. I'm not aware as yet of any such proof, hence my question.

Answer 1

Hint: show that $a_n-1\leq 1+\frac12+\frac14+\cdots+\frac1{2^{n-1}}$

Answer 2

By the binomial theorem $$1<a_n=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+...+\frac{1}{n!}\left(1-\frac{1}{n}\right)...<$$ $$<2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}<2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}<...$$ Can you end it now?

Answer 3

I would also use Binomial theorem but would estimate a bit differently using a telescoping sum:

\begin{eqnarray*} \left(1+\frac{1}{n}\right)^{n} & = & 1+1 +\sum_{k=2}^n\frac{n(n-1)\cdots (n-k+1)}{n^k}\cdot\frac{1}{k!}\\ & < & 2 +\sum_{k=2}^n\frac{1}{(k-1)k}\\ & = & 2 +\sum_{k=2}^n\left( \frac{1}{k-1}-\frac{1}{k}\right)\\ & & 2 + 1- \frac{1}{n} = 3-\frac{1}{n} \end{eqnarray*}

Answer 4

Hint:

1)$\binom{n}{k}\frac{1}{n^k} \le \frac {1}{k!}, k \in \mathbb{N}$.

2)$(1+ \frac{1}{n})^n =$

$\sum_{k=0}^{n} \binom{n}{k}(\frac{1}{n})^k \le \sum_{k=0}^{n}\frac{1}{k!}$

3) Upper bound:

$\sum_{k=0}^{n} \frac{1}{k!} \le 1+ \sum_{k=0}^{n}\frac{1}{2^k} < 3$.

Answer 5

Without using $\,b_n\,$ you can use of course what the others have written!

There is nothing to add.

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Given: $\,a_n\,$ monotone increasing and $\,b_n\,$ monotone decreasing and $\,a_n<b_n\,$ for all $\,n>0$

Proof: $~2= a_1 \leq a_n \leq \lim\limits_{n\to\infty} a_n = \lim\limits_{n\to\infty} b_n \leq b_n \leq b_1 = 4$

Therefore $\{a_n\}$ is bounded. That's really the shortest way.

Answer 6

If $x \geqslant y > 0$, and $n$ is a positive integer, then $$ x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1}) \geqslant n(x - y)y^{n-1}. $$ Therefore, for $n > 1$, \begin{align*} a_n - a_{n-1} & = \left(1+\frac{1}{n}\right)^n \! - \left(1+\frac{1}{n-1}\right)^{n-1} \\ & = \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \left[ \left(1+\frac{1}{n-1}\right)^{n-1} \!\! - \left(1+\frac{1}{n}\right)^{n-1}\right] \\ & \leqslant \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{1}{n^2}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{a_n}{(n+1)^2}, \end{align*} whence $$ a_n \leqslant a_{n-1}\left(1 - \frac{1}{(n+1)^2}\right)^{-1} \quad (n > 1). $$ By induction on $n$, $$ a_n \leqslant 2\left(1 - \frac{1}{9}\right)^{-1}\!\! \left(1 - \frac{1}{16}\right)^{-1}\!\!\cdots \left(1 - \frac{1}{(n+1)^2}\right)^{-1} \quad (n > 1). $$ Writing $c_n = (n+1)^{-2}$ and $s_n = c_2+c_3+\cdots+c_n$ ($n > 1$), we have $$ s_n < \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)} < \frac{1}{2} \quad (n > 1). $$ By the Weierstrass Product Inequality (the very simple proof by induction on $n$ is given here, but it could be left as an exercise), $$ (1 - c_2)(1 - c_3)\cdots(1 - c_n) \geqslant 1 - s_n \quad (n > 1). $$ So we have, finally, $$ a_n \leqslant 2(1 - c_2)^{-1}(1 - c_3)^{-1}\cdots(1 - c_n)^{-1} \leqslant 2(1 - s_n)^{-1} < 4. $$