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

  • Thread starter Thread starter Oggy
  • Start date Start date
  • Tags Tags
    Proof
Oggy
Messages
26
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
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!
 
Corrected now, thanks :) (It's (n-i)!)
 
Thanks for the correction, but I still can't obtain the correct answer for \sum_{i=0}^{1} A_i. Puzzling...
 
In the sum i goes from 0 to n. And it's (-1)^(n-i). Sorry for the mistakes.
 
Last edited:
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...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top