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

AI Thread Summary
The discussion centers on proving the factorial identity \(\frac{2 i}{(2 i + 1)!} = \frac{1}{(2 i)!} - \frac{1}{(2 i + 1)!}\). Participants suggest manipulating the left-hand side (LHS) by expressing \((2i + 1)!\) in terms of \((2i)!\) to simplify the equation. A key suggestion involves finding a common denominator on the right-hand side (RHS) to facilitate the proof. Additionally, some participants recommend adding and subtracting 1 in the numerator to combine terms effectively. The conversation emphasizes the importance of strategic manipulation to achieve the desired equality.
Paul245
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\frac{2 i}{(2 i + 1)!} = \frac{1}{(2 i)!} - \frac{1}{(2 i + 1)!}

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:
\frac{2 i}{(2 i + 1)!} = \frac{1}{(2 i)!} - \frac{1}{(2 i + 1)!}

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!)
 
\frac{2i}{(2i+1)!}=\frac{2i}{(2i+1)(2i)! }
=\frac{2i}{2i(2i)! + (2i)! }
= \frac{2i}{2i(2i)! + 2i (2i - 1)! }
= \frac{1}{(2i)! + (2i - 1)! }

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

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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