       Re: a tricky limit

• To: mathgroup at smc.vnet.net
• Subject: [mg15525] Re: [mg15327] a tricky limit
• From: Daniel Lichtblau <danl>
• Date: Mon, 18 Jan 1999 23:47:25 -0500
• References: <199901080915.EAA03988@smc.vnet.net.>
• Sender: owner-wri-mathgroup at wolfram.com

```Arnold Knopfmacher wrote:
>
> I wish to obtain a numerical estimate (say 8 decimal digits) of the
> limit as x  tends to 1 from below of  the function
> h[x]=(Product[(1-fm[x]/(m+1)),{m,2,Infinity}])/(1-x) where
> fm[x]=x^(m-m/d) and d is the smallest divisor of m that is greater than
> 1. The problem is that when I replace Infinity by say 1000 as the upper
> limit  of the product, the function blows up near 1. Visual inspection
> of the graph of h[x] for 0<x<0.9 say, suggests that the limit should
> have a value around 2.1. Can anyone help?
>
> Thanks
> Arnold Knopfmacher
> Dept of Computational and Applied Math Witwatersrand University
> South Africa

It may be a seriuos challenge to get that much precision. If you work
with the log of this product you can use power series expansions to
approximate the limit. First I'll show some crude bounds.

Define

htop[x_] := Product[(1-upper[x,m]/(m+1)),{m,2,Infinity}]/(1-x)
upper[x_,m_] := x^(m-1)

hbot[x_] := Product[(1-lower[x,m]/(m+1)),{m,2,Infinity}]/(1-x)
lower[x_,m_] := x^Ceiling[m/2]

Note that htop majorizes your h, while hbot lies below it. I will use
them to give some explicit bounds below.

It is clear that Log[h[x]] <
Sum[Log[1-x^(k-1)/(k+1), {k,2,Infinity}] - Log[1-x]

The right side can be rewritten as

Sum[2/(k*(k+2))*x^k, {k,1,Infinity}] -
Sum[x^(j*k)/(j*(k+2)^j), {j,2,Infinity}, {k,1,Infinity}]

We can now rewrite the second sum (I'll need to use a meta-notation
here) as

Sum[x^k / ((d+2)^(k/d)*(k/d)), d divides k, d<k, and k from 2 to
Infinity]

So we have log[h[x]] < 2*x/3 + Sum[
(2/(k*(k+2)) - Sum[(d/k)*(d+2)^(-k/d), {d<k and d divides k}]) * x^k,
{k,2,Infinity}]

This approaches 2/3 + Sum[2/(k*(k+2)), {k,2,Infinity}] - positive

where positive is strictly less than

Sum[1/k * 3^(-k), {k,2,Infinity}] +
Sum[1/k * 4^(-k), {k,2,Infinity}] +
1/2 * Sum[1/(k+2)^2, {k,3,Infinity}]

If you evaluate the mess you find that Log[h] (in the limit as x -> 1
from below, that is) is less than 1.28, and h is thus bounded above
by 3.6.

Note that minor modification of all this can be used to show that the
limit in question actually exists.

I'll skip the details of a crude overestimate. It is quite similar to
the underestimate above except one works with lower[x,m] instead of
upper[x,m]. The details are now a bit trickier, but not by much. I
obtained the following:

Log[h[x]] > 2/3*x + Sum[1/(4*k^2+2*k), {k,2,Infinity}]
- Sum[d/k * (2*d+1)^(-k/d) * x^k, d < k and d divides k,
{k,2,Infinity}]
- Sum[d/k * (2*d)^(-k/d) * x^k, d < k and d divides k,
{k,4,Infinity}]

A bit of work indicates this is majorized by

2/3*x + Sum[1/(4*k^2+2*k)*x^k, {k,2,Infinity}]
- 1/18*x^2 - 1/81*x^3 -
Sum[1/k^2*Log[2,k]*x^k, {k,4,Infinity}]

Plugging in x->1 we get

2/3 + Sum[1/(4*k^2+2*k), {k,2,Infinity}] - 1/18 - 1/81 -
Sum[1/k^2*Log[2,k], {k,4,Infinity}]

which gives me

97/162 + (5 - 3*Log)/6 -
(-9*Log - 4*Log - 36*Derivative[Zeta])/(36*Log)

or about -0.1875. Hence Log[h] > -0.1875 and so h > 0.83
(approximately). Which means that your guess of 2.1 is in the correct
ballpark.

f[m_][x_] := x^(m-m/Divisors[m][]) h[n_][x_] :=
Product[1-f[k][x]/(k+1), {k,2,n}]

and then go with

logh[x_,n_] := Normal[Series[PowerExpand[Log[
h[2*n][x]/(1-x)]], {x,0,n}]]

then you can approximate Log[h] by

In:= logh[x,16] /. x->1

1049092804861383602377884900513626863 Out=
-------------------------------------
1261370601365496207234698649600000000

In:= N[%]
Out= 0.831709

In:= Exp[%]
Out= 2.29724

I've done something similar to about 4000 terms using numeric
approximations to avoid hefty rational arithmetic. I get h
approximately equal to 2.292, and I am not very confident even about
that last decimal place. I believe that the error after n terms is
roughly O(1/n) though I do not have a full proof as yet. One way to
gauge this is to use degrees 10, 100, 1000 vs, say, 11,101,1001 and
each time see wheether one more decimal place has stabilized (I have
not done this myself).

I find it mildly comforting, albeit odd, that my upper and lower bounds
average to something quite close, that is, 2.21.

Daniel Lichtblau
Wolfram Research

```

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• From: "Arnold Knopfmacher" <arnoldk@gauss.cam.wits.ac.za>
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