# Prove that $a_n = \{\left(1+\frac{1}{n}\right)^n\}$ is bounded sequence, $n\in\mathbb{N}$

How to prove the following:

$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $n\in\mathbb{N}$

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You have at least two typos. –  Andres Caicedo Mar 27 '13 at 4:53
I think this statement is not true since $a_n> n^n$. –  Wayson Kong Mar 27 '13 at 4:53
I guess you meant $(1+\frac{1}{n})^n$. It can be proved using the inequality of arithmetic and geometric means –  Long Mar 27 '13 at 5:02
The best idea is not to prove it. –  Marc van Leeuwen Mar 27 '13 at 7:59

Let me denote $S_n=\{(1+\frac{1}{n})^n\}$ \begin{align*} \bigg(1+\frac{1}{n}\bigg)^n &=1+ {_nC_1} \frac{1}{n}+ {_nC_2} \bigg(\frac{1} {n}\bigg)^2+..........+{_nC_n} \bigg(\frac{1}{n}\bigg)^n\\ &=1+ n \frac{1}{n}+\frac{n(n-1)}{2!} \bigg(\frac{1}{n}\bigg)^2+ \frac{n(n-1)(n-2)}{2!} \bigg(\frac{1}{n}\bigg)^3+..........+\frac{n(n-1)(n-2)....(n-(n-1))}{n!}\bigg(\frac{1}{n}\bigg)^n\\ &=1+ 1+ \frac{(1-\frac{1}{n})}{2!} + \frac{(1-\frac{1}{n})(1-\frac{2}{n})} {3!}+..........+\frac{(1-\frac{1}{n})(1-\frac{2}{n}).....(n-\frac{n-1}{n})} {n!}\\ &<1+ 1+ \frac{1}{2!} + \frac{1}{3!}+..........+\frac{1}{n!}\\ &<1+ 1+ \frac{1}{2} + \frac{1} {2^2}+..........+\frac{1}{2^{n-1}}\\ &=1+\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}}\\ &=1+2\bigg(1-\frac{1}{2^n}\bigg)\\ &=3-\frac{1}{2^{n-1}}\\ &<3\\ \end{align*}

$S_n$ is bounded by 3.

Also note that $S_n>2$.

Here I'm not writing the reason in every step i hope u will understand.

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This can be shown very quickly with the inequality $\log (1+u) \leq u$. Indeed, $$\left(1+\frac{1}{n}\right)^n = \exp\left(n\log(1+\frac{1}{n})\right) \leq\exp\left(n\times\frac{1}{n}\right)$$ This gives you $e$ as an upper bound. Actually, this is the best possible bound since it is also the limit of the sequence.

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As Weson Jiang notes, $a_n = \left(n + \frac{1}{n}\right)^n > n^n$, and $n^n\to\infty$ as $n\to\infty$, so $a_n\to\infty$ as well.

You might have wanted to show that $\left\{b_n\right\}_{n\in\Bbb{N}} = \left\{\left(1 + \frac{1}{n}\right)^n\right\}_{n\in\Bbb{N}}$ is a bounded sequence. The binomial theorem implies $\left(1 + \frac{1}{n}\right)^n = \sum_{k = 0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{1}{n^k}$, so \begin{align*} \left(1 + \frac{1}{n}\right)^n &= \sum_{k = 0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{1}{n^k}\\ &=\frac{1}{1} + n\frac{1}{n} + \begin{pmatrix}n\\2\end{pmatrix}\frac{1}{n^2} + \ldots + \frac{1}{n^n}\\ &\leq 1 + 1 + \frac{n^2}{2!}\frac{1}{n^2} + \frac{n^3}{3!}\frac{1}{n^3} + \ldots + \frac{n^n}{n!}\frac{1}{n^n}\\ &< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} + \ldots\\ &< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^n} + \ldots\\ &= 1 + \sum_{k = 0}^{\infty}\frac{1}{2^k}\\ &= 1 + 2 = 3, \end{align*} so the sequence is bounded above by $3$.

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Below is the same answer as the one by Ju'x, in slightly different language. We want to show that $\left(1+\dfrac{1}{n}\right)^n \lt e$, or equivalently that $1+\frac{1}{n}\lt e^{1/n}$. The series expansion of $e^{1/n}$ is a sum of positive terms, the first two of which are $1$ and $\dfrac{1}{n}$.

The downside is that this approach is not available if we intend to define $e$ as the limit of $\left(1+\dfrac{1}{n}\right)^n$.

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Thanks Andre (+1) –  Babak S. Mar 27 '13 at 7:31
+1: I like that way of thinking. –  Siméon Mar 27 '13 at 8:47

There was a comment that you can use $\text{AM} \ge \text{GM}$ to prove the boundedness of this.

Here is a proof. As a side effect, we also prove the convergence.

First we show that $x_n = \left(1 + \frac{1}{n}\right)^n$ is monotonically increasing. We prove this, using $\text{AM} \ge \text{GM}$.

We have that, by taking $n$ copies of $\left(1 + \frac{1}{n}\right)$ and one copy of $1$ that,

$$\frac{\left(1 + \frac{1}{n}\right) + \dots + \left(1 + \frac{1}{n}\right) + 1}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$

i.e.

$$\frac{n+2}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$

and so

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

Now we use $\text{AM} \ge \text{GM}$ (again!).

We take $n$ copies of $1$ and one copy of $\frac{1}{2}$ to get

$$\frac{n + \frac{1}{2}}{n+1} \ge \sqrt[n+1]{\frac{1}{2}}$$

i.e.

$$2^{\frac{1}{n+1}} \ge \frac{2n+2}{2n+1}$$ i.e

$$2 \ge \left(1 + \frac{1}{2n+1}\right)^{n+1}$$

And so $$4 \ge \left(1 + \frac{1}{2n+1}\right)^{2n+1}$$

Since the sequence is monotonically increasing, this bound applies to the whole sequence.

(Side effect: Since the sequence is monotonic, and bounded, it is convergent)

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