One Plus Reciprocal to the Nth

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Theorem

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \paren {1 + \dfrac 1 n}^n$.

Then $\sequence {x_n}$ converges to a limit as $n$ increases without bound.


Proof

First we show that $\sequence {x_n}$ is increasing.

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

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:

$A_n = \dfrac {\paren {n - 1} \paren {1 + \dfrac 1 {n - 1} } + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
$G_n = \paren {1 + \dfrac 1 {n - 1} }^{\dfrac {n - 1} n}$


By Cauchy's Mean Theorem‎:

$G_n \le A_n$

Thus:

$\paren {1 + \dfrac 1 {n - 1} }^{\frac {n - 1} n} \le 1 + \dfrac 1 n$

and so:

$x_{n - 1} = \paren {1 + \dfrac 1 {n - 1} }^{n - 1} \le \paren {1 + \dfrac 1 n}^n = x_n$

Hence $\sequence {x_n}$ is increasing.


Next, we show that $\sequence {x_n}$ is bounded above.

Using the Binomial Theorem:

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

So $\sequence {x_n}$ is bounded above by $3$.


From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x_n}$ converges to a limit.


Also see

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

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