Why is gamma(z+1) = z * gamma(z)?

[shnaiwer]

hi to all ...
it is just an attempt to understand ...
the problem was to prove that
gamma ( z+ 1) = z * gamma(z) using the integral definition of gamma function ...
when i used the integration by parts i get the following :

gamma ( z+ 1 ) = ( x^z) * e^(-x) (0 ,∞) + ⌠ z x^(z-1) e^(-x ) dx
where the limits of integration ( 0 ,∞)

the integration already gives z * gamma (z) , then ( x^z) * e^(-x) │( 0 ,∞) = 0

i don't have the evidense that this term is zero ...

have u ?

 

[Gib Z]

What does the " | " actually denote? What values do you get when you sub in 0 and take the limit as x goes to infinity ? It should be quite straight forward. Remember exponential terms dominate algebraic terms in limits.

 

[shnaiwer]

What does the " | " actually denote? What values do you get when you sub in 0 and take the limit as x goes to infinity ? It should be quite straight forward. Remember exponential terms dominate algebraic terms in limits.

thank u ... " | " denote nothing other than ( when or at ) , we can neglect it..
when we sub in zero we get zero , since
( 0^z) * e^(-0) = 0 * 1 = 0

but the problem still hold when we try that limit as x goes to infinity , where it gives
( ∞ / ∞ ) or ( 0 / 0) if we try l'Hôpital's rule , in fact i want to understand or get the answer of the question :
why do exponential terms dominate algebraic terms in limits ?
i think that you put the exact description of what my question originally is , when you said : Remember exponential terms dominate algebraic terms in limits

 

[Dick]

Take the log of L=x^z*e^(-x), log(L)=z*log(x)-x. x grows faster than z*log(x). Use l'Hopital to prove it. So the limit of the log(L) goes to negative infinity. That means L goes to zero.