let r be real number ; for seq being Real_Sequence st 0 <= r & ( for n being Element of NAT holds seq . n = 1 / ((n * n) + r) ) holds
lim seq = 0
let seq be Real_Sequence; ( 0 <= r & ( for n being Element of NAT holds seq . n = 1 / ((n * n) + r) ) implies lim seq = 0 )
assume that
A1:
0 <= r
and
A2:
for n being Element of NAT holds seq . n = 1 / ((n * 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 )consider k1 being
Element of
NAT such that A4:
p " < k1
by Th10;
assume A5:
0 < p
;
ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs ((seq . m) - 0) < pthen A6:
k1 > 0
by A4;
then
k1 >= 1
+ 0
by NAT_1:13;
then
k1 <= k1 * k1
by XREAL_1:153;
then A7:
k1 + r <= (k1 * k1) + r
by XREAL_1:8;
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
n * n <= m * m
by XREAL_1:68;
then A9:
(n * n) + r <= (m * m) + r
by XREAL_1:8;
(p ") + 0 < k1 + r
by A1, A4, XREAL_1:10;
then
(p ") + 0 < (k1 * k1) + r
by A7, XXREAL_0:2;
then
1
/ ((k1 * k1) + r) < 1
/ (p ")
by A5, XREAL_1:78;
then A10:
1
/ ((k1 * k1) + r) < 1
* ((p ") ")
by XCMPLX_0:def 9;
0 < n ^2
by A6, SQUARE_1:74;
then
1
/ ((m * m) + r) <= 1
/ ((n * n) + r)
by A1, A9, XREAL_1:120;
then A11:
1
/ ((m * m) + r) < p
by A10, XXREAL_0:2;
(
seq . m = 1
/ ((m * m) + r) &
0 <= m * m )
by A2;
hence
abs ((seq . m) - 0) < p
by A1, A11, ABSVALUE:def 1;
verum end;
seq is convergent
by A1, A2, Th47;
hence
lim seq = 0
by A3, SEQ_2:def 7; verum