Is This Factorial Identity True and How Can It Be Proven?

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    Factorial Identity
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Discussion Overview

The discussion centers around the verification and proof of a factorial identity involving the terms \(\frac{2i}{(2i + 1)!}\) and \(\frac{1}{(2i)!} - \frac{1}{(2i + 1)!}\). Participants explore various approaches to manipulate the left-hand side (LHS) to match the right-hand side (RHS), focusing on mathematical reasoning and algebraic manipulation.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents the identity and requests assistance in transforming the LHS to the RHS.
  • Another participant suggests using the factorial relationship \((2i + 1)! = (2i + 1)(2i!)\) to simplify the expression.
  • Several participants provide algebraic manipulations, showing steps to rewrite the LHS in terms of factorials and combining terms.
  • One participant critiques the approach taken by others, suggesting a different method involving finding a common denominator on the RHS.
  • Another participant proposes an alternative method of adding and subtracting 1 in the numerator to facilitate the transformation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to prove the identity, with multiple competing methods and suggestions presented throughout the discussion.

Contextual Notes

Some steps in the algebraic manipulations remain unresolved, and there are dependencies on specific assumptions about factorial properties that are not explicitly stated.

Paul245
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[itex]\frac{2 i}{(2 i + 1)!}[/itex] = [itex]\frac{1}{(2 i)!}[/itex] - [itex]\frac{1}{(2 i + 1)!}[/itex]

Could anybody please show what it is that needs to be done on LHS to get to RHS in this identity.
 
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Paul245 said:
[itex]\frac{2 i}{(2 i + 1)!}[/itex] = [itex]\frac{1}{(2 i)!}[/itex] - [itex]\frac{1}{(2 i + 1)!}[/itex]

Could anybody please show what it is that needs to be done on LHS to get to RHS in this identity.

(2i + 1)! = (2i+1)(2i!)
 
[itex]\frac{2i}{(2i+1)!}[/itex]=[itex]\frac{2i}{(2i+1)(2i)! }[/itex]
=[itex]\frac{2i}{2i(2i)! + (2i)! }[/itex]
= [itex]\frac{2i}{2i(2i)! + 2i (2i - 1)! }[/itex]
= [itex]\frac{1}{(2i)! + (2i - 1)! }[/itex]

and now what?
 
Paul245 said:
[itex]\frac{2i}{(2i+1)!}[/itex]=[itex]\frac{2i}{(2i+1)(2i)! }[/itex]
=[itex]\frac{2i}{2i(2i)! + (2i)! }[/itex]
= [itex]\frac{2i}{2i(2i)! + 2i (2i - 1)! }[/itex]
= [itex]\frac{1}{(2i)! + (2i - 1)! }[/itex]

and now what?

Nah, you've gone at that the wrong way.

Start with the question, and use coolul007's hint to get a common denominator on the right hand side.
 
Another way to go about: Add and subtract 1 in the numerator and club 2i and 1.
 
thanks acabus, all
 

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