Understanding the Sum of Factorial Products and Its Application in Counting Permutations

Factorial products and their sums are fascinating topics in mathematics, especially when applied in combinatorial contexts such as counting permutations. This article explores the sum of a specific series involving factorial products, presenting both mathematical proofs and a counting argument to demonstrate the solution.

Introduction

Consider the sum of the series:

$$ S 1 cdot 1! 2 cdot 2! 3 cdot 3! ldots n cdot n! $$

Our goal is to find a simplified formula for this sum. We will explore two approaches: a mathematical identity and a counting argument.

Mathematical Derivation Using Factorial Identities

The first approach involves expressing the general term in a different form:

$$ k cdot k! k! cdot k k cdot 1! - (k-1)! $$

Using this identity, we can rewrite the sum ( S ) as:

$$ S (2! - 1!) (3! - 2!) (4! - 3!) ldots (n! - (n-1)!) $$

This is a telescoping series where most terms cancel out:

$$ S n cdot 1! - 1! $$

Since ( 1! 1 ), we have:

$$ S n cdot 1! - 1 $$

Therefore, the final result for the sum is:

$$ S n cdot 1! - 1 $$

The answer can be boxed as follows:

$$ boxed{n cdot 1! - 1} $$

Counting Permutations: A Combinatorial Proof

Another way to prove the same result is through a counting proof. Consider the set ([n] {1, 2, ldots, n}).

For each ( k in [n] ), let ( g_{nk} ) count the permutations ( sigma ) of ([n 1]) for which ( k 1 ) is the largest non-fixed point. This means:

( sigma(k 1) eq k 1 ), ( sigma(r) r ) for all ( r ) such that ( k 1 leq r leq n ).

For such a permutation ( sigma ), we know that ( sigma(k 1) ) must be one of the elements in ([k]). Additionally, ( sigma ) maps ( [k] ) bijectively to ([k] setminus {sigma(k 1)} ). Therefore, the number of such permutations ( g_{nk} ) is:

$$ g_{nk} k cdot k! $$

Summing up all ( g_{nk} ) for ( k ) from 1 to ( n ), we count all permutations of ([n 1]) except the identity permutation, which has no non-fixed points:

$$ sum_{k1}^{n} g_{nk} (n cdot 1! - 1!) $$

Thus, the result is:

$$ sum_{k1}^{n} g_{nk} n cdot 1! - 1 $$

This confirms the same result using a counting argument.

Expression of the Product Series

Let's consider the product series:

$$ 11! cdot 22! cdot 33! cdot ldots cdot n^n $$

This can be re-written using the factorial identity:

$$ frac{2!}{1!} cdot frac{3!}{2!} cdot frac{4!}{3!} cdot ldots cdot frac{n!}{(n-1)!} cdot frac{(n 1)!}{n!} $$

The series simplifies as follows:

$$ frac{2! cdot 3! cdot 4! cdot ldots cdot n!}{1! cdot 2! cdot 3! cdot ldots cdot (n-1)!} - (1! 2! 3! ldots n!) $$

This simplifies further to:

$$ frac{n!}{1!} - (1! 2! 3! ldots n!) $$

This is equivalent to:

$$ n! - (1! 2! 3! ldots n!) $$

Therefore, the sum can be boxed as follows:

$$ boxed{n! - (1! 2! 3! ldots n!)} $$

Conclusion

In summary, we have explored two methods to find the sum of the series ( S 1 cdot 1! 2 cdot 2! 3 cdot 3! ldots n cdot n! ). The mathematical identity approach and the counting argument both lead to the same result:

$$ S n cdot 1! - 1 $$

This article has provided a clear understanding of the sum of factorial products and its application in counting permutations, offering insights for further exploration in combinatorial mathematics.