stirlinglem9

Description: ( ( BN ) - ( B( N + 1 ) ) ) is expressed as a limit of a series. This result will be used both to prove that B is decreasing and to prove that B is bounded (below). It will follow that B converges in the reals. (Contributed by Glauco Siliprandi, 29-Jun-2017)

	Ref	Expression
Hypotheses	stirlinglem9.1	$⊢ A = (n \in ℕ ⟼ \frac{n!}{\sqrt{2 n} {(\frac{n}{e})}^{n}})$
	stirlinglem9.2	$⊢ B = (n \in ℕ ⟼ \log A (n))$
	stirlinglem9.3	$⊢ J = (n \in ℕ ⟼ (\frac{1 + 2 n}{2}) \log (\frac{n + 1}{n}) - 1)$
	stirlinglem9.4	$⊢ K = (k \in ℕ ⟼ (\frac{1}{2 k + 1}) {(\frac{1}{2 \cdot N + 1})}^{2 k})$
Assertion	stirlinglem9	$⊢ N \in ℕ \to seq 1 (+ K) ⇝ (B (N) - B (N + 1))$

Proof

Step	Hyp	Ref	Expression
1		stirlinglem9.1	$⊢ A = (n \in ℕ ⟼ \frac{n!}{\sqrt{2 n} {(\frac{n}{e})}^{n}})$
2		stirlinglem9.2	$⊢ B = (n \in ℕ ⟼ \log A (n))$
3		stirlinglem9.3	$⊢ J = (n \in ℕ ⟼ (\frac{1 + 2 n}{2}) \log (\frac{n + 1}{n}) - 1)$
4		stirlinglem9.4	$⊢ K = (k \in ℕ ⟼ (\frac{1}{2 k + 1}) {(\frac{1}{2 \cdot N + 1})}^{2 k})$
5		eqid	$⊢ (k \in ℕ_{0} ⟼ 2 (\frac{1}{2 k + 1}) {(\frac{1}{2 \cdot N + 1})}^{2 k + 1}) = (k \in ℕ_{0} ⟼ 2 (\frac{1}{2 k + 1}) {(\frac{1}{2 \cdot N + 1})}^{2 k + 1})$
6	3 4 5	stirlinglem7	$⊢ N \in ℕ \to seq 1 (+ K) ⇝ J (N)$
7	1 2 3	stirlinglem4	$⊢ N \in ℕ \to B (N) - B (N + 1) = J (N)$
8	6 7	breqtrrd	$⊢ N \in ℕ \to seq 1 (+ K) ⇝ (B (N) - B (N + 1))$

Metamath Proof Explorer

Theorem stirlinglem9

Proof