Visual proof of $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + ... = 1$ I am trying to find a visual way of proving that $$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + ... = 1$$
I am familiar with algebraic methods of proving this formulas (such as using a telescoping sum) - this is not what I am looking for.
I was thinkng about each term as representing the areas of rectangles (see image) but I am unsure about how to use this to prove the above result.

 A: There exists a visual proof of this sum without the use of rectangles and areas. I saw this in Proofs Without Words. I give two explanations, first using coordinate geometry, second using similarity of right triangles.
Explanation using coordinate geometry :
Consider the lines intercepting the axes at $(n,0)$ and $(0,1)$ for $n=1,2,3,\ldots$

Equation of lines is
$$\frac{x}{n}+\frac{y}{1}=1$$
They make an intercept of $1-\dfrac{1}{n}$ on $x=1$ line. Hence length of the colored segments is
$$\left(1-\frac{1}{n+1}\right)-\left(1-\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$$
which are the terms in our sum. As $n$ becomes arbitrarily large, these lines approach the horizontal $y=1$ line. Clearly the sum of colored segments tends to unity.
Explanation using similarity of triangles :
We have right triangles with both sides on axes with legs $1$ and $n$. All the hypotenuse intersect with the colored vertical line to produce similar triangles with base $n-1$. Thus the similarity ratio is $\dfrac{n-1}{n}$ and the vertical leg is also of length $\dfrac{n-1}{n}\cdot 1=1-\dfrac{1}{n}$.
Same as above, each differently colored segment has length $\dfrac{1}{n(n+1)}$ and as number of triangles becomes arbitrarily large, our sum approaches value of $1$.
