#### To determine

**To sketch:** The curves
y=xn,0≤x≤1
for
n=0,1,2,3,⋯
on a common screen and show that the areas between successive curves
∑n=1∞1n(n+1)=1
.

#### Answer

The sketch of the curves
y=xn,0≤x≤1
for
n=0,1,2,3,⋯
is shown in the Figure 1.

#### Explanation

Use online graphing calculator and draw the curves
y=xn,0≤x≤1
for
n=0,1,2,3,⋯
is shown in the Figure 1.

From Figure 1, it is observed that the areas of the curve become unit square.

The area between
y=xn−1
and
y=xn
for
0≤x≤1
is computed as follows,

∫01(xn−1−xn)dx=[xn−1+1n−1+1−xn+1n+1]01=[xnn−xn+1n+1]01=1n−1n+1=(n+1)−nn(n+1)

That is,
∫01(xn−1−xn)dx=1n(n+1)
.

As
n→∞
, the sum of the areas between the successive curves approaches the area of the unit square.

That is,
∑n=1∞1n(n+1)=1
.

Hence, the required proof is obtained.