let r be real number ; for seq being Real_Sequence st 0 <= r & ( for n being Element of NAT holds seq . n = 1 / (n + r) ) holds
lim seq = 0
let seq be Real_Sequence; ( 0 <= r & ( for n being Element of NAT holds seq . n = 1 / (n + r) ) implies lim seq = 0 )
assume that
A1:
0 <= r
and
A2:
for n being Element of NAT holds seq . n = 1 / (n + r)
; lim seq = 0
A3:
now let p be
real number ;
( 0 < p implies ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs ((seq . m) - 0 ) < p )assume A4:
0 < p
;
ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs ((seq . m) - 0 ) < pconsider k1 being
Element of
NAT such that A5:
p " < k1
by Th10;
(p " ) + 0 < k1 + r
by A1, A5, XREAL_1:10;
then
1
/ (k1 + r) < 1
/ (p " )
by A4, XREAL_1:78;
then A6:
1
/ (k1 + r) < 1
* ((p " ) " )
by XCMPLX_0:def 9;
take n =
k1;
for m being Element of NAT st n <= m holds
abs ((seq . m) - 0 ) < plet m be
Element of
NAT ;
( n <= m implies abs ((seq . m) - 0 ) < p )assume
n <= m
;
abs ((seq . m) - 0 ) < pthen A7:
n + r <= m + r
by XREAL_1:8;
0 < p "
by A4;
then
1
/ (m + r) <= 1
/ (n + r)
by A1, A5, A7, XREAL_1:120;
then A8:
1
/ (m + r) < p
by A6, XXREAL_0:2;
(
seq . m = 1
/ (m + r) &
0 <= m )
by A2, NAT_1:2;
hence
abs ((seq . m) - 0 ) < p
by A1, A8, ABSVALUE:def 1;
verum end;
seq is convergent
by A1, A2, Th43;
hence
lim seq = 0
by A3, SEQ_2:def 7; verum