Formula of the n-th term of a sequence

[fmaurica]

Hi,
I'm trying to determine the formula of the n-th term of this sequence :
u_n = n*u_{n-1} - 1
I think I already dealt with that kind of sequence back to my high school years but I cannot remember anymore. I feel like a change of variable would be helpful.
Thanks.

 

[stapel]

I'm trying to determine the formula of the n-th term of this sequence :
u_n = n*u_{n-1} - 1

Just to clarify, which of the following is the correct formulation?

. . . . .\(\displaystyle \mbox{a) }\, u_n\, =\, n\cdot u_{n\, -\, 1}\, -\, 1\)

. . . . .\(\displaystyle \mbox{b) }\, u_n\, =\, n\left( u_{n\, -\, 1}\, -\, 1\right)\)

Thank you!

 

[HallsofIvy]

I would start by calculating a few value. You don't give "u_0" which, of course, you need as a starting point. Let's call it "a". Then u_1= a- 1, u_2= 2(a- 1)- 1= 2a- 3, u_3= 3(2a+ 3)- 1= 6a- 11, a_4= 4(6a- 11)-1= 24a- 45. It seems clear that the first term is "n! a". Now what about that constant term? Taking a= 0, since the constant term does not depend upon a, we would get -1, -3 , -11, -45, 5(-45)- 1= -226, 6(-226)- 1=-1357, etc.

 

[fmaurica]

Thanks for your quick answer.
"a) u_{n =} n.u_n-1 - 1" is the correct formulation.

Based on HallsofIvy's reply, we have :
u_n = n!u₀ + v_n where v_n is the sequence below
v₀ = 0 and v_n =n.v_n-1 - 1 which actually is... u_n (with the precision that the first term is zero). Aren't we going into a vicious loop ?

PS: the correct sequence for the constant term is -1, -3, -10, -41, -206, -1031, etc