# Prove that $\sum_{i=0}^{2n}(-1)^i\frac{1}{i!(2n-i)!} = 0$ for any $n>2$

I've been trying to solve this problem lately, but I have been unable to do it. I want to prove that

$$\sum_{i=0}^{2n}(-1)^i\frac{1}{i!(2n-i)!} = 0$$

For any $n>2$. We can generalize the problem changing $2n$ to $a$, I don't mind, although I am focusing on the even numbers, thus the reason for using $2n$.

I have no trouble in showing that the sum is 0 for $n = \infty$, but I am interested in finding a solution for any given $n$. I have tried using induction, but since the $n$ variable is also in the denominator I cannot cleanly let the term out of the expression.

If you multiply with $(2n)!$, then with use of Binomial theorem you get

$$\sum_{i=0}^{2n}(-1)^i\frac{(2n)!}{i!(2n-i)!} = \sum_{i=0}^{2n}(-1)^i{2n\choose i} = (1-1)^{2n} =0$$

• Thanks for the answer! I was so close-minded in trying to solve it by induction that I could not imagine making use of Binomial Theorem. – Alex Martínez Ascensión Oct 25 '17 at 21:52

Surely, this question has been answered here before. I am only giving an alternative probabilistic/combinatorial proof.

Let $C$ be the collection of all subsets of $[k]:=\{1,2,\ldots,k\}$, where $k$ is a fixed positive integer. Denote by $E$ the subcollection of $C$ consisting of sets with even number of elements, and write $O$ for the complement of $E$ in $C$. Then, there exists an involution $i:C\to C$ such that $i$ maps $E$ bijectively onto $O$ and, likewise, $i$ maps $O$ bijectively onto $E$. Such a map is given by $i(s):=s\cup \{1\}$ if $1\notin s$, whilst $i(s):=s\setminus\{1\}$ if $1\in s$, where $s\in C$ is arbitrary. (In other words, $i(s)=s\Delta \{1\}$, where $\Delta$ is the symmetric difference binary operator.) This shows that $|E|=|O|$. Additionally, if $C$ is given a discrete uniform probability measure, then this translates into the statement below:

The probability of picking a set in $C$ with an even number of elements is exactly $\frac{1}{2}$, which is the same as the probability of picking a set in $C$ with an odd number of elements.

To finish the proof, note that the probability of picking an element in $C$ of size $j\in\{0,1,2,\ldots,k\}$ is $$\frac{1}{k!}\,\binom{k}{j}=\frac{1}{j!\,(k-j)!}\,.$$ Finally, $k=2n$ is a particular case of this result.