logarithm reciprocal limit
Henryk Trappmann
sequence:
Define a_n recursively in the following way (for any b>0):
a_1 = 1 and for n>1:
a_n = 1/(1-b^n) * sum from m=1 to n-1 over a_m * (n over m) * (1-b)^(n-
m) * b^n
For ASCII handicapped people here the typeset formula:
http://math.eretrandre.org/cgi-bin/mimetex.cgi?a_n%20=%20\frac{1}{1-b^n}\sum_{m=1}^{n-1}%20a_m%20\left(n\\m\right)%20(1-b)^{n-m}%20b^m
My conjecture is that
a_n -> (b-1)/ln(b)
Have anyone heard about such a formula, or can prove or disprove its
convergence?
It sounds too elementary for not being treated somewhere already.
William Elliot
> Define a_n recursively in the following way (for any b>0):
>
> a_1 = 1 and for n>1:
> a_n = 1/(1-b^n) * sum from m=1 to n-1 over a_m * (n over m)
> * (1-b)^(n-m) * b^n
What's (n over m)? Is it n/m, the Jacobi symbol or something else?
Is the sum from 1 to n-1 or from 1 to (n-1) over a_m?
No, that's confusion of variables. What does over a_m mean? 1/a_m?
What's a_n if b = 1?
> My conjecture is that
> a_n -> (b-1)/ln(b)
Factoring b^n from the sum, let's look at
lim(n->oo) b^n / (1 - b^n) = lim(n->oo) 1/(b^-n - 1)
= 0 if b in (0,1)
= -1 if b in (1,oo).
Thus if the sum approaches a finite value, a_n -> 0 for b in (0,1)
----
[Mod note: judging by the URL he posted, the "n over m" is probably
a binomial coefficient?]
Henryk Trappmann
> What's (n over m)? Is it n/m, the Jacobi symbol or something else?
> Is the sum from 1 to n-1 or from 1 to (n-1) over a_m?
> No, that's confusion of variables. What does over a_m mean? 1/a_m?
Ok, here the latex set formula:
a_n = \frac{1}{1-b^n}\sum_{m=1}^{n-1} a_m \left(n\\m\right) (1-b)^{n-
m} b^m
Yes, (n over m) means the binomial coefficient. "over a_m ..." was
meant to take the sum "over" the rest of the line.
> What's a_n if b = 1?
That's true this case has to be excluded. However if the conjecture is
true we can continue the function (in variable b) to b=1.
[Mod note: later on I got the following:
Oh I see my typo the last power is not to n but to m.
Here the latex typeset formula
a_n = \frac{1}{1-b^n}\sum_{m=1}^{n-1} a_m \left(n\\m\right) (1-b)^{n-
m} b^m
]
Valentin Fadeev
Using the formula for summing by parts: \sum_{m=1}^{n-1}u \Delta v =
uv|_{m=1}^n -\sum_{m=1}^{n-1}E[v] \Delta u, where E[f(x)]=f(x+1).
In this case u=a_m; \Delta v = \left(n\\m\right) (1-b)^{n-m} b^m.
Therefore, \Delta u = a_{m+1}-a_m, v=(1+1-b)^n=1 (binomial formula),
E[1]=1.
u(n)=a_n, v(n)=b^n
u(1)=a_1=1, v(1)=nb
\sum_{m=1}^{n-1}E[v] \Delta u = \sum_{m=1}^{n-1}(a_{m+1}-a_m) = a_n-1
Finally (omitting the external factor (1-b^n)^{-1} for a while for
clarity):
\sum_{m=1}^{n-1} a_m \left(n\\m\right) (1-b)^{n-m} b^m = a_n b^n-a_1 n
b-(a_n-1).
Now this gives an equation for a_n:
a_n=\frac{nb-1}{2(b^n-1)} (n>=2)
Hence, as n\to\infty, for b\in[0,1) the sequence behaves like bn/2;
for b\in(1,\infty) it tends to 0.
Valentin Fadeev
with boundary terms. The corrected derivation in LaTex can be viewed
here:
Henryk Trappmann
Your formula $\frac{1-(1-b)^n - nb(1-b)^{n-1}}{1-b^n}$ can not be true,
it does not give the correct values;
The values of a_n with b=2 are for n=1,2,...
1, 4/3, 10/7, 152/105, 314/217, 940/651, 5678/3937, 1447504/1003935,...
yours:
0, -4/3, 4/7, -8/15, 8/31, -4/21, 12/127, -16/255, 16/511
I think you can not apply summation by parts if the v depends on two
variables, namely n and m.
In the meantime I found a non-recursive simple formula:
a_n = (b-1)\sum_{k=1}^{n} \binom{n}{k} \frac{k (-1)^k}{1-b^k}
However it doesnt trigger anything familiar in my mind.
So, does anybody know why or whether
b_n=\sum_{k=1}^n \binom{n}{k} \frac{k (-1)^k}{1-b^k}
tends to 1/\ln(b)?
Here is picture of the numerical behavior of b_n for b=2:
http://math.eretrandre.org/tetrationforum/attachment.php?aid=719
It oscillates with decreasing amplitude and increasing period around
1/ln(2)=1.442695... (the y-axis numbers are however messed up by Sage)
Roland Franzius
>
> So, does anybody know why or whether
> b_n=\sum_{k=1}^n \binom{n}{k} \frac{k (-1)^k}{1-b^k}
> tends to 1/\ln(b)?
a_n =sum_{k=1}^n k binomial{n}{k} (-1)^k(1-b^k)
=sum_{k=1}^n k binomial{n}{k} (-x)^k |_x->1
-sum_{k=1}^n k binomial{n}{k} (-x)^k |_x->b
=(x d/dx sum_{k=0}^n binomial{n}{k} (-x)^k) |_x->1
-(x d/dx sum_{k=0}^n binomial{n}{k} (-x)^k) |_x->b
= (x d/dx (1-x)^n) |_b ^1
= n b (1-b)^n
But this expression seems to be not identical with
a(1,b_)=1
a(n_,b_):=a(n,b)=
Simplify[(1-b)^n/(1-b^n) Sum[Binomial[n,k]a[k,m] (b/ (1-b))^m) ]]
which has an irreducible polynomial in b with small alternating
symmetrical coefficents as numerator and a product of monomials of type
(1+b^2 +b^4 .. ) as denominators.
--
Roland Franzius
Henryk Trappmann
> Henryk Trappmann schrieb:
>>
>> So, does anybody know why or whether
>> b_n=\sum_{k=1}^n \binom{n}{k} \frac{k (-1)^k}{1-b^k}
>> tends to 1/\ln(b)?
>
>
>
> a_n =sum_{k=1}^n k binomial{n}{k} (-1)^k(1-b^k)
No, in my formula above there is a fraction \frac{k (-1)^k}{1-b^k}
without LaTeX: k * (-1)^k / ( 1 - b^k)
while in your formula it is a product (!) even omitting k,
thats something completely different.
For clarity the whole formula in ASCII:
b_n=sum(from k=1 to n) binomial(n over k) * k * (-1)^k/(1-b^k)
> a(1,b_)=1
> a(n_,b_):=a(n,b)=
> Simplify[(1-b)^n/(1-b^n) Sum[Binomial[n,k]a[k,m] (b/ (1-b))^m) ]]
The formula for a_n (which looks correct in the way you gave it) differs
by just a multiplicative constant from my above given b_n:
a_n = (b-1)*b_n
Henryk Trappmann
On 07/20/2010 03:00 PM, Henryk Trappmann wrote:
> My conjecture is that
> a_n -> (b-1)/ln(b)
I settled now a proof for that, including lots more background:
http://arxiv.org/abs/1008.1409