# Thread: A strange way to write 0

1. ## A strange way to write 0

Let $K \leq 3$ an integer. Prove that $$\underset{i=0}{\overset{K}{\sum}}\frac{\left(-1\right)^{i}}{i!}\frac{\left(K-i-1\right)^{2}}{\left(K-i\right)!}=0.$$

2. ## Re: A strange way to write 0

You can't prove it- it isn't true! Just calculate the K= 0 or K= 1 cases to see that .

3. ## Re: A strange way to write 0

It is true if you reverse the inequality. If $\displaystyle K\ge 3$, it holds. Here is a rough outline for a proof:

For $\displaystyle K\ge 3$:

\displaystyle \begin{align*}\sum_{i=0}^K\dfrac{(-1)^i(K-i-1)^2}{i!(K-i)!} & = \sum_{i=0}^K \dfrac{(-1)^i(K^2-2iK-2K+i^2+2i+1)}{i!(K-i)!} \\ & = \sum_{i=0}^K \dfrac{(-1)^i(K-1)^2}{i!(K-i)!} + \sum_{i=0}^K \dfrac{(-1)^i(i(i-1)+i(3-2K))}{i!(K-i)!} \\ & = \dfrac{(K-1)^2}{K!}\sum_{i=0}^K (-1)^i \dfrac{K!}{i!(K-i)!} + \dfrac{(-1)^0(0(-1)+0(3-2K))}{0!(K-0)!} + \sum_{i=1}^K\dfrac{(-1)^i(i(i-1)+i(3-2K))}{i!(K-i)!}\end{align*}

\displaystyle \begin{align*} & = \dfrac{(K-1)^2}{K!}\sum_{i=0}^K\binom{K}{i}(-1)^i + (3-2K)\sum_{i=1}^K\dfrac{(-1)^i}{(i-1)!(K-i)!} + \sum_{i=1}^K \dfrac{(-1)^i (i-1)}{(i-1)!(K-i)!} \\ & = \dfrac{(K-1)^2}{K!}(1-1)^K + (3-2K)\sum_{i=0}^{K-1}\dfrac{(-1)^{i+1}}{i!(K-1-i)!} + \dfrac{(-1)^1(1-1)}{(1-1)!(K-1)!} + \sum_{i=2}^K \dfrac{(-1)^i}{(i-2)!(K-i)!} \\ & = 0 + \dfrac{2K-3}{(K-1)!}\sum_{i=0}^{K-1}(-1)^i\dfrac{(K-1)!}{i!(K-1-i)!} + 0 + \dfrac{1}{(K-2)!}\sum_{i=0}^{K-2}(-1)^{i+2}\dfrac{(K-2)!}{i!(K-2-i)!} \\ & = \dfrac{2K-3}{(K-1)!}\sum_{i=0}^{K-1}\binom{K-1}{i}(-1)^i + \dfrac{1}{(K-2)!}\sum_{i=0}^{K-2}\binom{K-2}{i}(-1)^i \\ & = \dfrac{2K-3}{(K-1)!}(1-1)^{K-1} + \dfrac{1}{(K-2)!}(1-1)^{K-2} = 0+0 = 0\end{align*}

Edit: The crux of the proof involves trying to make each term resemble the binomial formula for $\displaystyle (1-1)^a$ for some power $\displaystyle a$.