reserve a,b,c,d for positive Real,
  m,u,w,x,y,z for Real,
  n,k for Nat,
  s,s1 for Real_Sequence;

theorem
  (for n st n>=1 holds s.n=sqrt(n) & s.0=0) implies for n st n>=1 holds
  Partial_Sums(s).n<1/6*(4*n+3)*sqrt(n)
proof
  defpred X[Nat] means Partial_Sums(s).$1<1/6*(4*$1+3)*sqrt($1);
  assume
A1: for n st n>=1 holds s.n=sqrt(n) & s.0=0;
A2: for n be Nat st n>=1 & X[n] holds X[n+1]
  proof
    let n be Nat;
    assume that
A3: n>=1 and
A4: Partial_Sums(s).n<1/6*(4*n+3)*sqrt(n);
    n+1>=1+1 by A3,XREAL_1:7;
    then n+1>=1 by XXREAL_0:2;
    then
A5: s.(n+1)=sqrt(n+1) by A1;
    0+(16*n^2+8*n)<1+(16*n^2+8*n) by XREAL_1:8;
    then (16*n+8)*n<1*(16*n^2+8*n+1);
    then (4*(4*n+1)+4)/((4*n+1)^2)<1/n by A3,XREAL_1:106;
    then (4*(4*n+1))/((4*n+1)*(4*n+1))+4/((4*n+1)^2)<1/n by XCMPLX_1:62;
    then 4/(4*n+1)+4/((4*n+1)^2)<1/n by XCMPLX_1:91;
    then 1+(4/(4*n+1)+4/((4*n+1)^2))<1+1/n by XREAL_1:8;
    then 1+4/(4*n+1)+4/((4*n+1)^2)<1+1/n;
    then 1+2*2/(4*n+1)+(2/(4*n+1))^2<1+1/n by XCMPLX_1:76;
    then 1+2*(2/(4*n+1))+(2/(4*n+1))^2<1+1/n by XCMPLX_1:74;
    then (2/(4*n+1)+1)^2<(1+n*1)/n by A3,XCMPLX_1:113;
    then sqrt((2/(4*n+1)+1)^2)<sqrt((1+n)/n) by SQUARE_1:27;
    then 2/(4*n+1)+1<sqrt((1+n)/n) by SQUARE_1:22;
    then (2+(4*n+1)*1)/(4*n+1)<sqrt((1+n)/n) by XCMPLX_1:113;
    then
A6: (4*n+3)/(4*n+1)<sqrt(1+n)/sqrt(n) by SQUARE_1:30;
    sqrt(n)>0 by A3,SQUARE_1:25;
    then (1/6)*((4*n+3)*(sqrt(n)))<(1/6)*((sqrt(1+n))*(4*n+1)) by A6,XREAL_1:68
,102;
    then Partial_Sums(s).n<(1/6)*sqrt(1+n)*((4*n)+7)-sqrt(1+n) by A4,XXREAL_0:2
;
    then
A7: Partial_Sums(s).n+sqrt(1+n)<(1/6)*sqrt(1+n)*((4*n)+7)-sqrt(1+n)+sqrt(
    1+n) by XREAL_1:8;
    thus thesis by A5,A7,SERIES_1:def 1;
  end;
  Partial_Sums(s).(1+0) =Partial_Sums(s).0 + s.(1+0) by SERIES_1:def 1
    .=s.0 + s.1 by SERIES_1:def 1
    .=0+s.1 by A1
    .=1 by A1,SQUARE_1:18;
  then
A8: X[1];
  for n be Nat st n>=1 holds X[n] from NAT_1:sch 8(A8,A2);
  hence thesis;
end;
