Interesting probability!

Let $X_n$ be the sum of $n$ such random numbers, and let $F_n(x)$ be the cumulative distribution function of $X_n$ . Conditional probability considerations tell us that $F_n(x) \; =\; \int_0^x F_{n-1}(x-y)\,dy \hspace{2cm} n \ge 1\,,\,0 \le x \le 1$ (the formula becomes much more complicated when $x > 1$ but, happily, we don't need that). Since $F_1(x) = x$ for $0 \le x \le 1$ it is a simple induction to show that $F_n(x) \; = \; \frac{x^n}{n!} \hspace{2cm} n \ge 1\,,\, 0 \le x \le 1$ and hence $\sum_{n \ge 2} P_n(\tfrac12) \; = \; \sum_{n\ge 2} \frac{1}{n! 2^n} \; = \; e^{\frac12} - \tfrac32 = \boxed{0.1487212707}$

I want to do this problem Geometrically.

Consider the 3 dimensional case . There we just have to find $\displaystyle \frac{\text{volume of tetrahendron}}{\text{volume of cube}}$ .

Where the tetrahedron is formed by the intersection of the plane $x+y+z = \frac{1}{2}$ with the 3 coordinate axes and origin. The volume of tetrahedron is $\displaystyle \frac{1}{6}\cdot\frac{1}{2^{3}}$ this can be done using calculus or scalor tripple product or any other way.

Now we want to extend this concept to higher dimensions namely $\mathbb{R}^{n}$

The way to do that is using Simplexes and hypercubes .

So the probability $\displaystyle P_{n}(0.5)$ is simply $\displaystyle \frac{\text{Volume of n-Simplex}}{\text{Volume of hypercube}}$ .

Now the question is how to find the volume of a n-simplex formed by the intersection of $x_{1}+x_{2}+....x_{n} = \frac{1}{2}$ with the n-coordinate axes and origin and lying in the first hyperoctant or orthant.

Well let us see the case for 4th dimension

Let us use calculus to find the volume formed by the intersection $x_{1}+x_{2}+x_{3}+x_{4} = \frac{1}{2}$ with the 4 coordinate axes .

It would be $\large \int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}-x_{1}}\int_{0}^{\frac{1}{2}-x_{2}-x_{1}}\int_{0}^{\frac{1}{2}-x_{3}-x_{2}-x_{1}} dx_{4}dx_{3}dx_{2}dx_{1}$ .

The evaluation of these integrals is easy because all of them are of the form $\int (at+b)^{k} dt$ where $a$ and $b$ are constants and $k$ is a natural number.

Solving the above integral we get $\large \frac{1}{4!}\frac{1}{2^{4}}$ .

So we can generalize this concept into 5th dimension , 6th dimension , and in this way nth dimension .

So the volume of the n-Simplex formed by intersection of $x_{1}+x_{2}+....x_{n} = \frac{1}{2}$ with the n-coordinate axes would just be $\Large \frac{1}{n!}\frac{1}{2^{n}}$ .

This can also be done using Determinants. The volume of a n-simplex formed by intersection of $x_{1}+x_{2}+....x_{n} = \frac{1}{2}$ with the n-coordinate axes is given by $\frac{1}{n!}det(A)$ . Where $A$ is a $n\times n$ diagonal matrix all of whose diagonal elements are $\frac{1}{2}$ .

And the total sample space is nothing but the unit hypercube of nth dimension. So the volume of the unit hypercube of nth dimension is nothing but $1^{n} = 1$ .

So in general we can say:-

$\displaystyle P_{n}(0.5) = \frac{\frac{1}{n!}\frac{1}{2^{n}}}{1}$ .

So $\displaystyle \sum_{r=2}^{\infty} P_{n}(0.5) = \sum_{r=2}^{\infty} \frac{1}{n!}\frac{1}{2^{n}}$

Using the taylor series for $\displaystyle e^{x} = \sum_{r=0}^{\infty} \frac{x^{r}}{r!}$ .

We have $\displaystyle \sum_{r=2}^{\infty} \frac{1}{n!}\frac{1}{2^{n}} = e^{\frac{1}{2}} - \frac{3}{2}$ .

Note:- The generalization of volume of an n-simplex formed by the intersection of $\large x_{1}+x_{2}+....x_{n} = a$ with the n-coordinate axes and lying in the first hyperoctant or orthant is given by:-

$\large \int_{0}^{a}\int_{0}^{a-x_{n}}\int_{0}^{a-x_{n-1}-x_{n}}......\int_{0}^{a-x_{3}-......-x_{n-1}-x_{n}}\int_{0}^{a-x_{2}-x_{3}-......-x_{n-1}-x_{n}} dx_{1}dx_{2}dx_{3}......dx_{n-1}dx_{n} = \frac{a^{n}}{n!}$