Can You Find a Finite Expression for the Summation of Factorial Terms?

[Atran]

Hi, I've been trying to make another expression for,

m
$$\sum$$(k!) = f(m)
k=0

I did one expression that is,

f(m) = 2(3(4(5...((m-1)(m+1)+1)...+1)+1)+1)+1

For instance, f(4) = 2(3(4+1)+1)+1

Can you hint me on finding a finite expression for the above expression?
Thanks for help.

 

[Xitami]

http://oeis.org/search?q=1,2,4,10,34,154,874,5914

 

[Goongyae]

FYI your function f(x) can be extended to non-integer argument x as follows:

Since

s! = integral(0,infinity)[ x^s e^-x dx]

it follows that

f(s) = integral(0,infinity)[ (1+x+x^2+...+x^s) e^-x dx]
= integral(0,infinity)[ (1-x^(s+1))/(1-x) *e^-x dx]

Numerical integration (not sure how accurate) yields
f(-1.5)=-1.21027
f(-0.5) = 0.562187
f(0.5) = 1.44841
f(1.5) = 2.77775
f(2.5) = 6.1011
f(i) = 0.963332 + 0.751815i

 

[Xitami]

f0(n)= sum(i=0, n-1, i!)
f1(n)= my(r=1,s=1); for(i=1,n-1,s*=i;r+=s);r
f2(n)= my(r=n);forstep(i=n-2,1,-1,r=r*i+1);r

 

[CellCoree]

Sure, I would be happy to provide some insight on finding a finite expression for the summation of factorial terms. First, it is important to understand the concept of factorials. A factorial is represented by an exclamation point (!) and is defined as the product of all positive integers less than or equal to the given number. For example, 4! = 4 x 3 x 2 x 1 = 24.

Now, let's break down the given expression:

m
\sum(k!) = f(m)
k=0

This is a summation notation, where k is the index starting from 0 and going up to m. This means that we are adding up all the factorial terms from 0! to m!.

One way to approach finding a finite expression for this summation is by using the concept of binomial coefficients. These coefficients represent the number of ways to choose k objects from a set of n objects. It can be calculated using the formula:

nCk = n! / (k!(n-k)!)

For our expression, we can rewrite it as:

m
\sum(k!) = 0 + 1! + 2! + 3! + ... + m!
k=0

Using the binomial coefficient formula, we can rewrite each term as:

0 + (1C1 x 1!) + (2C1 x 2!) + (3C1 x 3!) + ... + (mC1 x m!)

Simplifying this further, we get:

0 + 1 + 2(1!) + 3(2!) + ... + m((m-1)!)

Now, we can see that each term is a multiple of the previous term, with the first term being 1. So, we can rewrite this expression as:

f(m) = 1 x 1! + 2 x 2! + 3 x 3! + ... + m x m!

This is a finite expression for the summation of factorial terms and can be further simplified if needed. I hope this helps and good luck with your further exploration!