Geometric argument showing that the sum of the series $\sum_{n = 1}^{\infty} n^{-1}(n + 1)^{-1}$ is $1$

Question

I can't see the relationship between this telescopic series and the area under the curve for functions $f(x)=x^n$.

Answer 1

Hint The area between $f_n(x)$ and $f_{n+1}(x)$ is $n^{-1}(n+1)^{-1}$.

Answer 2

Write out the partial sums of the series into a telescopic sum: $$ s_n = \sum_{i=1}^n \frac{1}{i} - \frac{1}{i+1}$$ from here we already see that $$ s_n = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n+1} = 1 - \frac{1}{n+1}$$ Taking $n\to \infty$ we have that $s_n \to 1$... However, the question explicitly talks about geometry. Lets find out the areas of the functions $f_n(x) = x^n$: $$\int _0^1 f_0(x)\mathrm d x = \int_0^1 \mathrm d x = 1$$ $$\int _0^1 f_1(x)\mathrm d x = \int_0^1 x \mathrm d x = \frac{1}{2}$$ $$\int _0^1 f_2(x)\mathrm d x = \int_0^1 x^2 \mathrm d x = \frac{1}{3}$$ $$\int _0^1 f_2(x)\mathrm d x = \int_0^1 x^3 \mathrm d x = \frac{1}{4}$$ $$\int _0^1 f_2(x)\mathrm d x = \int_0^1 x^4 \mathrm d x = \frac{1}{5}$$ Hmmmm... The first is the area under a straight line, the second is the area under a triangle, the third is the area under a parabola, and so on: in the general case $$ \int _0^1 f_n(x) \mathrm d x = \int_0^1 x^n\mathrm d x= \frac{1}{n+1}$$

Here, I have plotted some of the functions

Now note that the $n$th term of the sum is the difference between the areas under curve $n$ and curve $n+1$. Below you see the areas represented in the first 4 terms

Keep adding areas that become smaller and smaller and you will have a full square with side 1 and area 1 :)