Shifted Factorial Homework: Showing \sum_{n=0}^{\infty}

  • Thread starter Thread starter Ted123
  • Start date Start date
  • Tags Tags
    Factorial
Click For Summary

Homework Help Overview

The problem involves demonstrating the equality of a series involving shifted factorials, specifically \(\sum_{n=0}^{\infty} \frac{(a)_n(-1)_n}{(c)_n n!}\) and \(\frac{c-a}{c}\). The discussion centers around the definitions and values of the shifted factorial (Pochhammer symbol) and the implications of the terms in the series.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of the shifted factorial and the implications of the term \((-1)_n\) for \(n \geq 2\). There are questions about the values of \((-1)_0\), \((-1)_1\), \((a)_0\), and \((a)_1\), as well as the subsequent terms \((a)_2\) and \((a)_3\).

Discussion Status

The discussion is active, with participants providing clarifications on the values of the factorial terms. Some guidance has been offered regarding the definitions, but there is no explicit consensus on the overall approach to the series.

Contextual Notes

Participants are working within the constraints of the problem statement and exploring the definitions of the shifted factorials, which may lead to different interpretations of the series terms.

Ted123
Messages
428
Reaction score
0

Homework Statement



I've got to show [tex]\sum_{n=0}^{\infty} \frac{(a)_n(-1)_n}{(c)_n n!} = \frac{c-a}{c}[/tex]
where
[tex]\displaystyle (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)} = a(a+1)...(a+n-1)[/tex]
is the shifted factorial (Pochhammer symbol).

The Attempt at a Solution



I've been informed that [tex](-1)_n = 0\;\;\;\;\;\;\forall\;\;n\geq 2[/tex]
So the sum has only 2 terms for n=0 and n=1, but what do e.g. [tex](-1)_0\,,\,(-1)_1\,,\,(a)_0\,,\,(a)_1[/tex] equal?
 
Physics news on Phys.org
Hi Ted123! :smile:

look at the definition of (a)n

(a)0 obviously = 1 for all a (including (-1)0 = 1),

and (a)1 = … ? :wink:
 
tiny-tim said:
Hi Ted123! :smile:

look at the definition of (a)n

(a)0 obviously = 1 for all a (including (-1)0 = 1),

and (a)1 = … ? :wink:

So would (a)1 = a, and (-1)1 = -1 ?

So the 2 terms of the sum give [tex]1 - \frac{a}{c} = \frac{c-a}{c}[/tex]
Incidentally would (a)2 = a(a+1), (a)3 = a(a+1)(a+2) etc.?
 
Last edited:
yes yes yes yes and yes :smile:
 

Similar threads

Replies
5
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K