let s be Real_Sequence; :: thesis: ( ( for n being Nat holds s . n <> 0 ) & ex m being Nat st
for n being Nat st n >= m holds
((abs s) . (n + 1)) / ((abs s) . n) >= 1 implies not s is summable )
assume A1: for n being Nat holds s . n <> 0 ; :: thesis: ( for m being Nat ex n being Nat st
( n >= m & not ((abs s) . (n + 1)) / ((abs s) . n) >= 1 ) or not s is summable )
given m being Nat such that A2: for n being Nat st n >= m holds
((abs s) . (n + 1)) / ((abs s) . n) >= 1 ; :: thesis: not s is summable

A3: now :: thesis: for n being Nat st n >= m holds
|.(s . n).| >= |.(s . m).|

defpred S₁[ Nat] means |.(s . (m + $1)).| >= |.(s . m).|;
A4: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A5: |.(s . (m + k)).| >= |.(s . m).| ; :: thesis: S₁[k + 1]
((abs s) . ((m + k) + 1)) / ((abs s) . (m + k)) >= 1 by A2, NAT_1:11;
then |.(s . ((m + k) + 1)).| / ((abs s) . (m + k)) >= 1 by SEQ_1:12;
then A6: |.(s . ((m + k) + 1)).| / |.(s . (m + k)).| >= 1 by SEQ_1:12;
s . (m + k) <> 0 by A1;
then |.(s . (m + k)).| > 0 by COMPLEX1:47;
then |.(s . ((m + k) + 1)).| >= |.(s . (m + k)).| by A6, XREAL_1:191;
hence S₁[k + 1] by A5, XXREAL_0:2; :: thesis: verum

end;

let n be Nat; :: thesis: ( n >= m implies |.(s . n).| >= |.(s . m).| )
assume n >= m ; :: thesis: |.(s . n).| >= |.(s . m).|
then consider k being Nat such that
A7: n = m + k by NAT_1:10;
reconsider k = k as Element of NAT by ORDINAL1:def 12;
A8: n = m + k by A7;
A9: S₁[ 0 ] ;
for k being Nat holds S₁[k] from NAT_1:sch 2(A9, A4);
hence |.(s . n).| >= |.(s . m).| by A8; :: thesis: verum

end;

s . m <> 0 by A1;
then |.(s . m).| > 0 by COMPLEX1:47;
then ( not s is convergent or lim s <> 0 ) by A3, Th38;
hence not s is summable by Th4; :: thesis: verum