How Do You Prove the Sum of This Complex Series Equals One?

[Oggy]

Let $$A_i = \frac{1}{n}\cdot \frac{(-1)^{n-i}}{i!\cdot(n-i)!} \int_{0}^{n} \frac{t(t-1)...(t-n)}{t-i}dt$$

I need to prove

$$\sum_{i=0}^{n} A_i = 1$$.

I tried tinkering with the equation but I'm really at a loss what to do with the integral. I'd appreciate any help.

 

[pizzasky]

Error?

Hi! Is there an error somewhere?

I tried evaluating $$\sum_{i=0}^{1} A_i$$ but my answer was 0, and not 1. Perhaps you can re-check the question?

All the best!

 

[Oggy]

Corrected now, thanks :) (It's (n-i)!)

 

[pizzasky]

Thanks for the correction, but I still can't obtain the correct answer for $$\sum_{i=0}^{1} A_i$$. Puzzling...

 

[Oggy]

In the sum i goes from 0 to n. And it's (-1)^(n-i). Sorry for the mistakes.

 

[pizzasky]

Well, the Binomial Series is probably involved... seeing the factorials and the term $$(-1)^{n-i}$$, but apart from that, I am not very sure how to proceed...