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e and in a mirror

Benoit Cloitre is a brilliant maths amateur when he wants.

1 Two mirrored sequences

Consider the sequences n and n to be defined as

 u1= 0, u2 = 1, un+2 = un+1 + v1= 0, v2 = 1, vn+2 = + vn
Then In general, for

 u1= x, u2 = y, un+2 = un+1 + Then

lim n   = x + By considering those two sequences, e and are mirrored! However, the speed of convergence are very different... Un converges very quickly (faster and faster) while V n converges in a logarthmic way...

2 Prove

Benoit Cloitre first offered a direct prove, but one can prefer Gery Huvent's prove, from Lille, which allow us to find a general method to prove those sequences.

2.1 For n

Calculating the first few terms 0,1,1, , , , suggest that This property can be prove by induction on p, since it's true for p = 1. Suppose that v2p = v2p+1 then and

 v2p+3 = + v2p+1 = * vp + v2p = vp = v2p+2
hence the result. We also notice that the sequence p satisfy which by induction give us Taking into account the classical equivalence (obtain with Wallis) we have Note We can find a limited development with Stirling's formula, to get 2.2 For n

Consider the function that generates the sequence n the given recurrent sequence can be written as or by differentiating, we get = 0

hence f is the solutiong to the differential equation which has the general solution taking into acount than u1 = 0 and u2 = 1 = we find Now we need to determine the development of the power serie of f so to find each term of the sequence. For this note that ( = v2x + v3x2 + ....is defined in 0 )
Then, we remember that if then using Cauchy's product of two series. Hence and  and more particularly Note If we did the same for the sequence n, the generating function is a solution to  which satisfy with 