let f be Real_Sequence; :: thesis: ( ( for n being Element of NAT holds f . n = log 2,(n ! ) ) implies for n being Element of NAT holds f . n = Sum seq_logn ,n )
assume A1:
for n being Element of NAT holds f . n = log 2,(n ! )
; :: thesis: for n being Element of NAT holds f . n = Sum seq_logn ,n
set g = seq_logn ;
defpred S1[ Element of NAT ] means f . $1 = Sum seq_logn ,$1;
A2:
S1[ 0 ]
A4:
for k being Element of NAT st S1[k] holds
S1[k + 1]
proof
let k be
Element of
NAT ;
:: thesis: ( S1[k] implies S1[k + 1] )
assume A5:
f . k = Sum seq_logn ,
k
;
:: thesis: S1[k + 1]
A6:
k ! > 0
by NEWTON:23;
f . (k + 1) =
log 2,
((k + 1) ! )
by A1
.=
log 2,
((k + 1) * (k ! ))
by NEWTON:21
.=
(log 2,(k + 1)) + (log 2,(k ! ))
by A6, POWER:61
.=
(log 2,(k + 1)) + (Sum seq_logn ,k)
by A1, A5
.=
(seq_logn . (k + 1)) + (Sum seq_logn ,k)
by Def2
.=
(seq_logn . (k + 1)) + ((Partial_Sums seq_logn ) . k)
by SERIES_1:def 6
.=
(Partial_Sums seq_logn ) . (k + 1)
by SERIES_1:def 1
.=
Sum seq_logn ,
(k + 1)
by SERIES_1:def 6
;
hence
S1[
k + 1]
;
:: thesis: verum
end;
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A2, A4);
hence
for n being Element of NAT holds f . n = Sum seq_logn ,n
; :: thesis: verum