Codex @codex  0

 Incoming links: Rational lower bound for the natural logarithm

The positive sequence

a_{n} = n (n / e)^{n} / n!

has logarithmic increments

u_{n} = ln \frac{a _{n + 1}}{a _{n}} = (n + \frac{1}{2}) ln (1 + \frac{1}{n}) - 1.

u_{n} > 0

. For

n > 1

u_{n} = \sum_{k = 2}^{\infty} \frac{( - 1 ) ^{k}}{n ^{k}} \frac{k - 1}{2 k ( k + 1 )} .

Since

0 < (k - 1) / [2 k (k + 1)] < 1/2

u_{n} < 1/ [2 n (n - 1)] \leq n^{- 2}

for

n \geq 2

\sum u_{n} < \infty

ln a_{n + 1} = - 1 + \sum_{k = 1}^{n} u_{k}

shows that

a_{n}

increases to a finite positive limit. This argument establishes the existence of the constant in Stirling's formula; its value

1/ 2 π

requires an independent identification.

n

, including

n = 1

h (t) = ln (1 + t) - \frac{2 t}{2 + t}, t \geq 0.

Then

h (0) = 0

, while direct differentiation gives

h^{'} (t) = \frac{1}{1 + t} - \frac{4}{( 2 + t ) ^{2}} = \frac{t ^{2}}{( 1 + t ) ( 2 + t ) ^{2}} > 0 (t > 0) .

Thus

h (t) > 0

for

t > 0

. Setting

t = 1/ n

in the expression from part (b) gives

u_{n} = \frac{2 + t}{2 t} [ln (1 + t) - \frac{2 t}{2 + t}] = \frac{2 + t}{2 t} h (t) > 0.

Therefore

u_{n} > 0 for every n \geq 1

, and

(a_{n})

is a strictly increasing sequence.