let s be Real_Sequence; :: thesis: ( s is summable implies for n being Element of NAT holds s ^\ n is summable )
defpred S₁[ Element of NAT ] means s ^\ $1 is summable ;
A1: for n being Element of NAT st S₁[n] holds
S₁[n + 1]

proof

let n be Element of NAT ; :: thesis: ( S₁[n] implies S₁[n + 1] )
reconsider s1 = NAT --> ((s ^\ n) . 0) as Real_Sequence ;
for k being Element of NAT holds s1 . k = (s ^\ n) . 0 by FUNCOP_1:13;
then A2: Partial_Sums ((s ^\ n) ^\ 1) = ((Partial_Sums (s ^\ n)) ^\ 1) - s1 by Th14;
assume s ^\ n is summable ; :: thesis: S₁[n + 1]
then Partial_Sums (s ^\ n) is convergent by Def2;
then ( s ^\ (n + 1) = (s ^\ n) ^\ 1 & Partial_Sums ((s ^\ n) ^\ 1) is convergent ) by A2, NAT_1:49, SEQ_2:25;
hence S₁[n + 1] by Def2; :: thesis: verum

end;

assume s is summable ; :: thesis: for n being Element of NAT holds s ^\ n is summable
then A3: S₁[ 0 ] by NAT_1:48;
thus for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A3, A1); :: thesis: verum