# Stirling's Formula/Proof 2/Lemma 2

## Theorem

The sequence $\left \langle {d_n} \right \rangle$ defined as:

$\displaystyle d_n = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n$

is decreasing.

## Proof

The proof strategy is to demonstrate that the sign of $d_n - d_{n+1}$ is positive.

 $\displaystyle d_n - d_{n+1}$ $=$ $\displaystyle \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \left({\ln \left({\left({n+1}\right)!}\right) - \left({n + 1 + \frac 1 2}\right) \ln \left({n + 1}\right) + n + 1}\right)$ $\displaystyle$ $=$ $\displaystyle - \ln \left({n+1}\right) - \left({n + \frac 1 2}\right) \ln n + \left({n + \frac 3 2}\right) \ln \left({n+1}\right) - 1$ (as $\ln \left({\left({n+1}\right)!}\right) = \ln \left({n+1}\right) + \ln \left({n!}\right)$) $\displaystyle$ $=$ $\displaystyle \left({n + \frac 1 2}\right) \ln \left({\frac {n+1} n}\right) - 1$ $(1):\quad$ $\displaystyle$ $=$ $\displaystyle \frac {2n + 1} 2 \ln \left({\frac {1 + \left({2n + 1}\right)^{-1} } {1 - \left({2n + 1}\right)^{-1} } }\right) - 1$

Let:

$f \left({x}\right) := \dfrac 1 {2 x} \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1$

for $\left\vert{x}\right\vert < 1$.

Then from Lemma 1:

$(2): \quad \displaystyle f \left({x}\right) = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}$

Thus $f \left({x}\right) > 0$ for $\left\vert{x}\right\vert < 1$.

Putting $x = \dfrac 1 {2n + 1}$ it can be seen that $(1)$ is $f \left({\dfrac 1 {2n + 1} }\right)$.

As $-1 < \dfrac 1 {2n + 1} < 1$ it can be seen that $(2)$ can be applied and so:

$\forall n \in \N: d_n - d_{n+1} \ge 0$

Thus $\left\langle{d_n}\right\rangle$ is a decreasing sequence.

$\blacksquare$