## AGM-related limits

Let there be two numbers a and b.
Take the AM and GM of those two call them respectively a_1 and b_1
Then take AM and GM of a_1 and b_1, and call them a_2 and b_2, … .
Continuing this way, if the limit of a_n and b_n as n goes to infinity is x and y
then find these two limit as n goes to infinity:
I) a_1 + a_2 + a_3 + … + a_nn·x
II) b_1 + b_2 + b_3 + … + b_nn·y

(Proposed by Sanchar Sharma)

Sachin points out x = y, the AGM of a and b, often denoted M(a,b). In fact, this is a well-known theorem (by Gauss I think).

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Let s₁(n) = a₁ + a₂ + a₃ +… + a_n – n·x, and s₁(n) = b₁ + b₂ + b₃ +… + b_n – n·y.

Then 2s₁(n) + s₂(n) =

= 2a₁ + 2a₂ + 2a₃ +… + 2a_n – 2n·x + b₁ + b₂ + b₃ +… + b_n – n·y =

= 2a₁ + (a₁ + b₁) + (a₂ + b₂) +… + (a_(n-1) + b_(n-1)) + b₁ + b₂ + b₃ +… b_n – n·(2x + y) =

= 2a₁ + (a₁ + a₂ + … + a_(n-1)) + (b₁ + b₂ +… + b_(n-1)) + b₁ + b₂ + b₃ +… b_n – n·(2x + y) =

= 2a₁ + (s₁(n) – a_n + n·x) + (s₂(n) – b_n + n·y) + s₂(n) + n·y – n·(2x + y) =

= s₁(n) + 2s₂(n) + 2a₁ – a_n – b_n – n·(x – y).

This gives

2s₁(n) + s₁(n) = s₁(n) + 2s₂(n) + 2a₁ – a_n – b_n – n·(x – y),

s₁(n) – s₁(n) = + 2a₁ – a_n – b_n – n·(x – y)

Using x = y = M(a,b) this gives lim[n→∞] (s₁(n) – s₁(n)) = 2a – 2M(a,b).

So at least the difference s₁(n) – s₂(n) has a limit. Notice that x = y is necessary for this to exist! If the limit of one more (linear) combination could be found the problem would be solved. Unfortunately I haven’t been able to find more (yet). – Someone?

Of course we still dont know if the limits of s₁(n) and s₂(n) exist. But either both or none do!

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Some numerical results.
Let a = x and b = 1 where x > 1. Then s₁(n) = Σ [k=1,n] (a_k – M(1,x)) and
s₂(n) = Σ [k=1,n] (b_k – M(1,x)) are funtions of x and n. Provided the limits exist,
Σ₁(x) ≡ lim [n→∞] Σ [k=1,n] (a_k – M(1,x)) and Σ₂(x) ≡ lim [n→∞] Σ [k=1,n] (b_k – M(1,x)) are functions of x.

Using Mathematica it seems like the sums for given x converge rapidly and using n = 10 gives almost the same value as larger n. For example (here I use the values for n = 10 and the four numbers are appr values of x, Σ₁(x), Σ₂(x), Σ₁(x)/Σ₂(x), 2a – 2M(a,b)).

{10, 7.0803, -4.41889, 11.4992, 11.4992}
{20, 16.4921, -9.17624, 25.6684, 25.6684}
{50, 47.353, -23.0023, 70.3553, 70.3553}
{70, 68.9654, -32.0087, 100.974, 100.974}
{100, 102.245, -45.3221, 147.567, 147.567}
{150, 159.184, -67.1502, 226.334, 226.334}

Notice that the last two values are equal confirming the earlier result.

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Graphs for Σ₁(x) and Σ₂(x), the other the ratio Σ₁(x)/Σ₂(x):

Upper curve: y = Σ₁(x) ≡ lim [n→∞] Σ [k=1,n] (a_k – M(1,x))
Lower curve: y = Σ₂(x) ≡ lim [n→∞] Σ [k=1,n] (b_k – M(1,x))

M(1,x) is known; it’s equal to a certain elliptic integral where x is a parameter.

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y = Σ₁(x)/Σ₂(x)

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Convergence proof.

For definiteness we assume that a ≥ b > 0. Then from the theory of the AGM it is known that

(*) a = a₁ ≥ a₂ ≥ … ≥ a_n ≥ a_(n+1) ≥ … ≥ M ≥ … ≥ b_(n+1) ≥ b_n ≥ … ≥ b₂ ≥ b₁ = b > 0 , where M ≡ M(a,b), the AGM of a and b. Also, by induction,

(**) a_(n+1) – b_(n+1) ≤ (a_n + b_n)/2 – b_n = (a_n – b_n)/2 ≤ … ≤ (a – b)/2^n.

From (*) and (**) 0 ≤ a_n – M ≤ a_n – b_n ≤ (a – b)/2^(n-1).

Since Σ [n=1,∞] (a – b)/2^(n-1) is convergent it follows that Σ [n=1,∞] (a_n – M) is also convergent.

Annonser