Given information: Given curves are y=xn  ,0≤x≤1 , for n=0,1,2,3,4......

Calculation:

The graph of the curves y=xn  ,0≤x≤1 , for n=0,1,2,3,4...... is shown below.

So to find the area between every two successive curves using integrals:

  A1,2=∫01(x0−x1)dxA3,2=∫01(x1−x2)dxDoing this for all curves.An.n−1=∫01(xn−xn+1)dx

Next to find the sum of all these areas which is the sum of integrals so:

  Sn=∑n=0∞∫01(xn−xn+1)dxSolve the integral :Sn=∑n=0∞∫01(xn−xn+1)dx.=∑n=0∞(xn+1n+1−xn+2n+2)|10=∑n=0∞[1n+1n+1−0n+1n+1−1n+2n+2+0n+2n+2]=∑n=0∞[1n+1−1n+2]=∑n=0∞[n+2−n−1(n+1)(n+2)].=∑n=0∞[1(n+1)(n+2)]Replace n+1 by nA=∑n=1∞[1n(n+1)].

The sum of all the areas is the Area of the square with side 1, therefore.

  A=1 .

Therefor ∑n=1∞[1n(n+1)]=1.

Hence proved.