let X be Banach_Algebra; :: thesis: ( (0. X) rExpSeq is norm_summable & Sum ((0. X) rExpSeq) = 1. X )
defpred S1[ set ] means (Partial_Sums ||.((0. X) rExpSeq).||) . $1 = 1;
A1: for n being Nat st S1[n] holds
S1[n + 1]
proof
let n be Nat; :: thesis: ( S1[n] implies S1[n + 1] )
assume A2: (Partial_Sums ||.((0. X) rExpSeq).||) . n = 1 ; :: thesis: S1[n + 1]
thus (Partial_Sums ||.((0. X) rExpSeq).||) . (n + 1) = 1 + (||.((0. X) rExpSeq).|| . (n + 1)) by A2, SERIES_1:def 1
.= 1 + ||.(((0. X) rExpSeq) . (n + 1)).|| by NORMSP_0:def 4
.= 1 + ||.(((1 / (n + 1)) * (0. X)) * (((0. X) rExpSeq) . n)).|| by Th14
.= 1 + ||.((0. X) * (((0. X) rExpSeq) . n)).|| by LOPBAN_3:38
.= 1 + ||.(0. X).|| by LOPBAN_3:38
.= 1 + 0 by LOPBAN_3:38
.= 1 ; :: thesis: verum
end;
(Partial_Sums ||.((0. X) rExpSeq).||) . 0 = ||.((0. X) rExpSeq).|| . 0 by SERIES_1:def 1
.= ||.(((0. X) rExpSeq) . 0).|| by NORMSP_0:def 4
.= ||.(1. X).|| by Th14
.= 1 by LOPBAN_3:38 ;
then A3: S1[ 0 ] ;
A4: for n being Nat holds S1[n] from NAT_1:sch 2(A3, A1);
 Partial_Sums ||.((0. X) rExpSeq).|| is constant by A4;
then A5: ||.((0. X) rExpSeq).|| is summable by SERIES_1:def 2;
defpred S2[ set ] means (Partial_Sums ((0. X) rExpSeq)) . $1 = 1. X;
A6: for n being Nat st S2[n] holds
S2[n + 1]
proof
let n be Nat; :: thesis: ( S2[n] implies S2[n + 1] )
assume (Partial_Sums ((0. X) rExpSeq)) . n = 1. X ; :: thesis: S2[n + 1]
hence (Partial_Sums ((0. X) rExpSeq)) . (n + 1) = (1. X) + (((0. X) rExpSeq) . (n + 1)) by BHSP_4:def 1
.= (1. X) + (((1 / (n + 1)) * (0. X)) * (((0. X) rExpSeq) . n)) by Th14
.= (1. X) + ((0. X) * (((0. X) rExpSeq) . n)) by LOPBAN_3:38
.= (1. X) + (0. X) by LOPBAN_3:38
.= 1. X by LOPBAN_3:38 ;
:: thesis: verum
end;
(Partial_Sums ((0. X) rExpSeq)) . 0 = ((0. X) rExpSeq) . 0 by BHSP_4:def 1
.= 1. X by Th14 ;
then A7: S2[ 0 ] ;
for n being Nat holds S2[n] from NAT_1:sch 2(A7, A6);
 hence ( (0. X) rExpSeq is norm_summable & Sum ((0. X) rExpSeq) = 1. X ) by A5, Th2; :: thesis: verum