Méthode des trapèzes :
A(k)=1/n*((k/n)²+(k+1)²/n²)*1/2
=1/(2n)*(1/n²)*(k²+k²+2k+1)
=1/(2n³)(k²+(k+1)²)
S(n)=A(0)+A(1)+...+A(n)
=(n+1)/(2n³)+ (∑k²)/(2n³)+(∑(k+1)²)/(2n³)
=1/(2n³)+1/(2n²)+(n(n+1)(2n+1))/(12n³)+(n+1)(n+2)(2n+3)/(12n³)
=1/(2n³)+1/(2n²)+(1+1/n)(2+1/n))/(12)+(1+1/n)(1+2/n)(2+3/n)/(12)
donc S(n) converge vers 0+0+(1*2)/12+(1*1*2)/12=1/3 si n→+∞
