let s be Real_Sequence; :: thesis: ( s is summable implies for n being Nat holds s ^\ n is summable )

defpred S_{1}[ Nat] means s ^\ $1 is summable ;

A1: for n being Nat st S_{1}[n] holds

S_{1}[n + 1]

then A3: S_{1}[ 0 ]
by NAT_1:47;

thus for n being Nat holds S_{1}[n]
from NAT_1:sch 2(A3, A1); :: thesis: verum

defpred S

A1: for n being Nat st S

S

proof

assume
s is summable
; :: thesis: for n being Nat holds s ^\ n is summable
let n be Nat; :: thesis: ( S_{1}[n] implies S_{1}[n + 1] )

set s1 = seq_const ((s ^\ n) . 0);

for k being Nat holds (seq_const ((s ^\ n) . 0)) . k = (s ^\ n) . 0 by SEQ_1:57;

then A2: Partial_Sums ((s ^\ n) ^\ 1) = ((Partial_Sums (s ^\ n)) ^\ 1) - (seq_const ((s ^\ n) . 0)) by Th11;

assume s ^\ n is summable ; :: thesis: S_{1}[n + 1]

then Partial_Sums (s ^\ n) is convergent ;

then ( s ^\ (n + 1) = (s ^\ n) ^\ 1 & Partial_Sums ((s ^\ n) ^\ 1) is convergent ) by A2, NAT_1:48;

hence S_{1}[n + 1]
by Def2; :: thesis: verum

end;set s1 = seq_const ((s ^\ n) . 0);

for k being Nat holds (seq_const ((s ^\ n) . 0)) . k = (s ^\ n) . 0 by SEQ_1:57;

then A2: Partial_Sums ((s ^\ n) ^\ 1) = ((Partial_Sums (s ^\ n)) ^\ 1) - (seq_const ((s ^\ n) . 0)) by Th11;

assume s ^\ n is summable ; :: thesis: S

then Partial_Sums (s ^\ n) is convergent ;

then ( s ^\ (n + 1) = (s ^\ n) ^\ 1 & Partial_Sums ((s ^\ n) ^\ 1) is convergent ) by A2, NAT_1:48;

hence S

then A3: S

thus for n being Nat holds S