Is $n! \sum_{i=0}^n{\frac{(-1)^i}{i!}}- (n-1)! \bigg[\sum_{i=0}^{n-2}{\frac{(-1)^i}{i!}}+...+\sum_{i=0}^{2}{\frac{(-1)^i}{i!}}\bigg]=(n-1)!$ true?

Question

I am in the middle of doing a problem and has this sort of expression. I have a feeling that the following equality holds: $$n! \sum_{i=0}^n{\frac{(-1)^i}{i!}}- (n-1)! \bigg[\sum_{i=0}^{n-2}{\frac{(-1)^i}{i!}}+\sum_{i=0}^{n-3}{\frac{(-1)^i}{i!}}+\sum_{i=0}^{n-4}{\frac{(-1)^i}{i!}}+...+\sum_{i=0}^{2}{\frac{(-1)^i}{i!}}\bigg]=(n-1)!$$

Does the above identity hold?

If it is, how we can show it?

Any help would be highly appreciated!

Answer 1

Indeed the identity is valid. Dividing OPs expression by $n!$ and putting the bracketed sum to the right we show

The following is valid for $n> 0$ \begin{align*} \sum_{i=0}^n\frac{(-1)^n}{i!}=\frac{1}{n}\left(1+\sum_{j=2}^{n-2}\sum_{i=0}^j\frac{(-1)^i}{i!}\right) \end{align*}

$$ $$

We obtain starting with the RHS \begin{align*} \frac{1}{n}&\left(1+\sum_{j=2}^{n-2}\sum_{i=0}^j\frac{(-1)^i}{i!}\right)\\ &=\frac{1}{n}\sum_{j=0}^{n-2}\sum_{i=0}^j\frac{(-1)^i}{i!}\tag{1}\\ &=\frac{1}{n}\sum_{i=0}^{n-2}\sum_{j=i}^{n-2}\frac{(-1)^i}{i!}\tag{2}\\ &=\frac{1}{n}\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}(n-i-1)\tag{3}\\ &=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}-\frac{1}{n}\sum_{i=1}^{n-2}\frac{(-1)^i}{(i-1)!}-\frac{1}{n}\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}\tag{4}\\ &=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}+\frac{1}{n}\sum_{i=0}^{n-3}\frac{(-1)^i}{i!}-\frac{1}{n}\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}\tag{5}\\ &=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}-\frac{1}{n}\frac{(-1)^{n-2}}{(n-2)!}\tag{6}\\ &=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}+\frac{(-1)^{n-1}}{n}(n-1)\tag{7}\\ &=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}+\frac{(-1)^{n-1}}{(n-1)!}+\frac{(-1)^n}{n!}\\ &=\sum_{i=0}^{n}\frac{(-1)^i}{i!}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{align*} and the claim is finished.

Comment:

• In (1) we observe that $$1=\sum_{j=0}^{1}\sum_{i=0}^j\frac{(-1)^i}{i!}$$

• In (2) we exchange the sums by noting that $$0\leq i\leq j\leq n-2$$

• In (3) we see the inner sum indexed with $j$ is $n-i-1$

• In (4) we split the sum according to $n-1-i$ and observe that in the middle sum the term with $i=0$ vanishes

• In (5) we shift the index of the middle sum by one and prepare so for telescoping

• In (6) we use telescoping

• In (7) we expand with $\frac{n-1}{n-1}$

Answer 2

Note that this identity would be equivalent to the following. If the original expression is true then the following holds:

$ \frac{\Gamma(n+1,-1)}{\Gamma(n)} = e+\sum_{k=2}^n {\frac{\Gamma(k+1,-1)}{\Gamma(k+1)}}$

I am not sure, however, if the original identity in the question is true.

Answer 3

You can rewrite your LHS this way: $$LHS = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{j=2}^{n-2} \sum_{i=0}^j \frac{(-1)^i}{i!}\right) \\ = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{j=0}^{n-2} \sum_{i=0}^j \frac{(-1)^i}{i!} - 1\right)$$ Swapping sums gives: $$LHS = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{i=0}^{n-2} \sum_{j=i}^{n-2} \frac{(-1)^i}{i!} - 1\right) \\ = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{i=0}^{n-2} (n-1-i) \frac{(-1)^i}{i!} - 1\right) \\ = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - \sum_{i=0}^{n-2} (n-1)!(n-1-i) \frac{(-1)^i}{i!} + (n-1)! \\ = (-1)^n (1 - n) + n! \sum_{i=0}^{n-2} \frac{(-1)^i}{i!} - \sum_{i=0}^{n-2} (n-1)!(n-1-i) \frac{(-1)^i}{i!} + (n-1)! \\ = (-1)^n (1 - n) + \sum_{i=0}^{n-2} \frac{(-1)^i}{i!} \left[ n! - (n-1)!(n-1-i) \right] + (n-1)! \\ = \underbrace{(-1)^n (1 - n) + \sum_{i=0}^{n-2} \frac{(-1)^i}{i!}(n-1)!(1+i)}_A + (n-1)!$$

Can you now show that $A = 0$ ?