Metamath Proof Explorer


Theorem esumgect

Description: "Send n to +oo " in an inequality with an extended sum. (Contributed by Thierry Arnoux, 24-May-2020)

Ref Expression
Hypotheses esumsup.1 φ B 0 +∞
esumsup.2 φ k A 0 +∞
esumgect.1 φ n * k = 1 n A B
Assertion esumgect φ * k A B

Proof

Step Hyp Ref Expression
1 esumsup.1 φ B 0 +∞
2 esumsup.2 φ k A 0 +∞
3 esumgect.1 φ n * k = 1 n A B
4 1 2 esumsup φ * k A = sup ran n * k = 1 n A * <
5 nfv n φ
6 nfcv _ n z
7 nfmpt1 _ n n * k = 1 n A
8 7 nfrn _ n ran n * k = 1 n A
9 6 8 nfel n z ran n * k = 1 n A
10 5 9 nfan n φ z ran n * k = 1 n A
11 simpr φ z ran n * k = 1 n A n z = * k = 1 n A z = * k = 1 n A
12 simplll φ z ran n * k = 1 n A n z = * k = 1 n A φ
13 simplr φ z ran n * k = 1 n A n z = * k = 1 n A n
14 12 13 3 syl2anc φ z ran n * k = 1 n A n z = * k = 1 n A * k = 1 n A B
15 11 14 eqbrtrd φ z ran n * k = 1 n A n z = * k = 1 n A z B
16 eqid n * k = 1 n A = n * k = 1 n A
17 esumex * k = 1 n A V
18 16 17 elrnmpti z ran n * k = 1 n A n z = * k = 1 n A
19 18 biimpi z ran n * k = 1 n A n z = * k = 1 n A
20 19 adantl φ z ran n * k = 1 n A n z = * k = 1 n A
21 10 15 20 r19.29af φ z ran n * k = 1 n A z B
22 21 ralrimiva φ z ran n * k = 1 n A z B
23 ovexd φ n 1 n V
24 simpll φ n k 1 n φ
25 fz1ssnn 1 n
26 25 a1i φ n 1 n
27 26 sselda φ n k 1 n k
28 24 27 2 syl2anc φ n k 1 n A 0 +∞
29 28 ralrimiva φ n k 1 n A 0 +∞
30 nfcv _ k 1 n
31 30 esumcl 1 n V k 1 n A 0 +∞ * k = 1 n A 0 +∞
32 23 29 31 syl2anc φ n * k = 1 n A 0 +∞
33 32 ralrimiva φ n * k = 1 n A 0 +∞
34 16 rnmptss n * k = 1 n A 0 +∞ ran n * k = 1 n A 0 +∞
35 33 34 syl φ ran n * k = 1 n A 0 +∞
36 iccssxr 0 +∞ *
37 35 36 sstrdi φ ran n * k = 1 n A *
38 36 1 sselid φ B *
39 supxrleub ran n * k = 1 n A * B * sup ran n * k = 1 n A * < B z ran n * k = 1 n A z B
40 37 38 39 syl2anc φ sup ran n * k = 1 n A * < B z ran n * k = 1 n A z B
41 22 40 mpbird φ sup ran n * k = 1 n A * < B
42 4 41 eqbrtrd φ * k A B