Factorial identity question

[Paul245]

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

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

 

[coolul007]

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

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!)

 

[Paul245]

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

and now what?

 

[acabus]

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

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.

 

[Sourabh N]

Another way to go about: Add and subtract 1 in the numerator and club 2i and 1.

 

[Paul245]

thanks acabus, all