Metamath Proof Explorer


Theorem faclimlem2

Description: Lemma for faclim . Show a limit for the inductive step. (Contributed by Scott Fenton, 15-Dec-2017)

Ref Expression
Assertion faclimlem2 M 0 seq 1 × n 1 + M n 1 + 1 n 1 + M + 1 n M + 1

Proof

Step Hyp Ref Expression
1 faclimlem1 M 0 seq 1 × n 1 + M n 1 + 1 n 1 + M + 1 n = m M + 1 m + 1 m + M + 1
2 nnuz = 1
3 1zzd M 0 1
4 1cnd M 0 1
5 nn0p1nn M 0 M + 1
6 5 nnzd M 0 M + 1
7 nnex V
8 7 mptex m m + 1 m + M + 1 V
9 8 a1i M 0 m m + 1 m + M + 1 V
10 oveq1 m = k m + 1 = k + 1
11 oveq1 m = k m + M + 1 = k + M + 1
12 10 11 oveq12d m = k m + 1 m + M + 1 = k + 1 k + M + 1
13 eqid m m + 1 m + M + 1 = m m + 1 m + M + 1
14 ovex k + 1 k + M + 1 V
15 12 13 14 fvmpt k m m + 1 m + M + 1 k = k + 1 k + M + 1
16 15 adantl M 0 k m m + 1 m + M + 1 k = k + 1 k + M + 1
17 2 3 4 6 9 16 divcnvlin M 0 m m + 1 m + M + 1 1
18 5 nncnd M 0 M + 1
19 7 mptex m M + 1 m + 1 m + M + 1 V
20 19 a1i M 0 m M + 1 m + 1 m + M + 1 V
21 peano2nn m m + 1
22 21 adantl M 0 m m + 1
23 22 nnred M 0 m m + 1
24 simpr M 0 m m
25 5 adantr M 0 m M + 1
26 24 25 nnaddcld M 0 m m + M + 1
27 23 26 nndivred M 0 m m + 1 m + M + 1
28 27 recnd M 0 m m + 1 m + M + 1
29 28 fmpttd M 0 m m + 1 m + M + 1 :
30 29 ffvelrnda M 0 k m m + 1 m + M + 1 k
31 12 oveq2d m = k M + 1 m + 1 m + M + 1 = M + 1 k + 1 k + M + 1
32 eqid m M + 1 m + 1 m + M + 1 = m M + 1 m + 1 m + M + 1
33 ovex M + 1 k + 1 k + M + 1 V
34 31 32 33 fvmpt k m M + 1 m + 1 m + M + 1 k = M + 1 k + 1 k + M + 1
35 15 oveq2d k M + 1 m m + 1 m + M + 1 k = M + 1 k + 1 k + M + 1
36 34 35 eqtr4d k m M + 1 m + 1 m + M + 1 k = M + 1 m m + 1 m + M + 1 k
37 36 adantl M 0 k m M + 1 m + 1 m + M + 1 k = M + 1 m m + 1 m + M + 1 k
38 2 3 17 18 20 30 37 climmulc2 M 0 m M + 1 m + 1 m + M + 1 M + 1 1
39 18 mulid1d M 0 M + 1 1 = M + 1
40 38 39 breqtrd M 0 m M + 1 m + 1 m + M + 1 M + 1
41 1 40 eqbrtrd M 0 seq 1 × n 1 + M n 1 + 1 n 1 + M + 1 n M + 1