More on Sequences — Question 8

Question 8 – Solution

Step 1: Evaluate the area exactly.

By the power rule, $$a_n=\left[\frac{x^{n+1}}{n+1}\right]_0^1=\frac1{n+1}\longrightarrow0.$$

Step 2: Prove monotonicity directly from the integrands.

For $0\le x\le1$, $0\le x^{n+1}\le x^n$, with strict inequality on $(0,1)$. Integrating over $[0,1]$ gives $0<a_{n+1}<a_n$, so the areas decrease.

Step 3: Explain the geometry rigorously.

For every fixed $x\in[0,1)$, $x^n\to0$; only the single endpoint $x=1$ remains at height $1$, and one point has zero width and contributes no area. For a quantitative version, fix $0<c<1$ and split the interval: $$\int_0^1x^n\,dx=\int_0^c x^n\,dx+\int_c^1x^n\,dx \le c^n+(1-c).$$ Given $\varepsilon>0$, first choose $c$ so that $1-c<\varepsilon/2$. Then choose $n$ so large that $c^n<\varepsilon/2$. The displayed bound gives $a_n<\varepsilon$, proving geometrically that the areas vanish.