let rseq be Real_Sequence; :: thesis: ( ( for n being Nat holds 0 <= rseq . n ) & rseq is summable & Sum rseq = 0 implies for n being Nat holds rseq . n = 0 )
assume that
A1: for n being Nat holds 0 <= rseq . n and
A2: rseq is summable and
A3: Sum rseq = 0 ; :: thesis: for n being Nat holds rseq . n = 0
A4: Partial_Sums rseq is bounded_above by A1, A2, SERIES_1:17;
A5: for n being Nat holds (Partial_Sums rseq) . n <= Sum rseq
proof
let n be Nat; :: thesis: (Partial_Sums rseq) . n <= Sum rseq
(Partial_Sums rseq) . n <= lim (Partial_Sums rseq) by A1, A4, SEQ_4:37, SERIES_1:16;
hence (Partial_Sums rseq) . n <= Sum rseq by SERIES_1:def 3; :: thesis: verum
end;
A6: Partial_Sums rseq is V50() by A1, SERIES_1:16;
now :: thesis: for n1 being Nat holds not rseq . n1 <> 0
given n1 being Nat such that A7: rseq . n1 <> 0 ; :: thesis: contradiction
A8: for n being Nat holds 0 <= (Partial_Sums rseq) . n
proof
let n be Nat; :: thesis: 0 <= (Partial_Sums rseq) . n
A9: ( n = n + 0 & (Partial_Sums rseq) . 0 = rseq . 0 ) by SERIES_1:def 1;
0 <= rseq . 0 by A1;
hence 0 <= (Partial_Sums rseq) . n by A6, A9, SEQM_3:5; :: thesis: verum
end;
(Partial_Sums rseq) . n1 > 0
proof
now :: thesis: ( ( n1 = 0 & (Partial_Sums rseq) . n1 > 0 ) or ( n1 <> 0 & (Partial_Sums rseq) . n1 > 0 ) )
per cases ( n1 = 0 or n1 <> 0 ) ;
case A10: n1 = 0 ; :: thesis: (Partial_Sums rseq) . n1 > 0
then (Partial_Sums rseq) . n1 = rseq . 0 by SERIES_1:def 1;
hence (Partial_Sums rseq) . n1 > 0 by A1, A7, A10; :: thesis: verum
end;
case A11: n1 <> 0 ; :: thesis: (Partial_Sums rseq) . n1 > 0
set nn = n1 - 1;
A12: ( (n1 - 1) + 1 = n1 & 0 <= rseq . n1 ) by A1;
A13: n1 in NAT by ORDINAL1:def 12;
0 <= n1 by NAT_1:2;
then 0 + 1 <= n1 by A11, INT_1:7, A13;
then A14: n1 - 1 in NAT by INT_1:5, A13;
then A15: (Partial_Sums rseq) . ((n1 - 1) + 1) = ((Partial_Sums rseq) . (n1 - 1)) + (rseq . ((n1 - 1) + 1)) by SERIES_1:def 1;
0 <= (Partial_Sums rseq) . (n1 - 1) by A8, A14;
hence (Partial_Sums rseq) . n1 > 0 by A7, A12, A15; :: thesis: verum
end;
end;
end;
hence (Partial_Sums rseq) . n1 > 0 ; :: thesis: verum
end;
hence contradiction by A3, A5; :: thesis: verum
end;
hence for n being Nat holds rseq . n = 0 ; :: thesis: verum