One Plus Reciprocal to the Nth

Theorem

Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = \left({1 + \dfrac 1 n}\right)^n$.

Then $\left \langle {x_n} \right \rangle$ converges to a limit.

Proof

• First we show that $\left \langle {x_n} \right \rangle$ is increasing.

Let $a_1 = a_2 = \cdots = a_{n-1} = 1 + \dfrac 1 {n-1}$ and let $a_n = 1$.

Let:

• $A_n$ be the arithmetic mean of $a_1 \ldots a_n$
• $G_n$ be the geometric mean of $a_1 \ldots a_n$

Thus:

• $\displaystyle A_n = \frac{\left({n - 1}\right) \left({1 + \frac 1 {n-1}}\right) + 1} n = \frac {n + 1} n = 1 + \frac 1 n$
• $\displaystyle G_n = \left({1 + \frac 1 {n-1}}\right)^{\frac {n-1}{n}}$

By Arithmetic Mean Never Less than Geometric Mean‎, $G_n \le A_n$.

Thus:

$\displaystyle \left({1 + \frac 1 {n-1}}\right)^{\frac {n-1}{n}} \le 1 + \frac 1 n$

... and so:

$\displaystyle x_{n-1} = \left({1 + \frac 1 {n-1}}\right)^{n-1} \le \left({1 + \frac 1 n}\right)^n = x_n$

Hence $\left \langle {x_n} \right \rangle$ is increasing.

• Next we show that $\left \langle {x_n} \right \rangle$ is bounded above.

We use the Binomial Theorem:

	\(\displaystyle \)	\(\displaystyle \left({1 + \frac 1 n}\right)^n\)	\(=\)	\(\displaystyle 1 + n \left({\frac 1 n}\right) + \frac {n \left({n-1}\right)} 2 \left({\frac 1 n}\right)^2 + \cdots + \left({\frac 1 n}\right)^n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + 1 + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n-1} n}\right) \frac 1 {n!}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle 1 + 1 + \frac 1 {2!} + \frac 1 {3!} + \cdots + \frac 1 {n!}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle 1 + 1 + \frac 1 2 + \frac 1 {2^2} + \cdots + \frac 1 {2^n}\)	\(\displaystyle \)	(because $2^{n-1} \le n!$)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \frac {1 - \left({\frac 1 2}\right)^n} {1 - \frac 1 2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + 2 \left({1 - \left({\frac 1 2}\right)^n}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle 3\)	\(\displaystyle \)

So $\left \langle {x_n} \right \rangle$ is bounded above by $3$.

• So, from the Monotone Convergence Theorem, $\left \langle {x_n} \right \rangle$ converges to a limit.

Note

Note that, although we have proved that this sequence converges to some limit less than 3 (and incidentally greater than 2), we have not at this stage determined exactly what this number actually is.

See Euler's number, where this sequence provides a definition of that number (one of several that are often used).

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.19$