Re: a tricky limit

• To: mathgroup at smc.vnet.net
• Subject: [mg15387] Re: a tricky limit
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Tue, 12 Jan 1999 03:14:50 -0500
• Organization: University of Western Australia
• References: <774kf9\$45s@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?

With the definitions,

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

series expansion of truncated products

In[3]:= h[10][x] + O[x, 1]
Out[3]=
-2        1153
----------- + ---- + O[-1 + x]
11 (-1 + x)   1155

In[4]:= h[20][x] + O[x, 1]
Out[4]=
-2        40020454
----------- + -------- + O[-1 + x]
21 (-1 + x)   33948915

indicates the leading (problem) term and the low-order approximations to
the limit you are after.  You can easily show that the leading
(problem) term, i.e.,

In[5]:= Product[m/(m + 1), {m, 2, n}] Out[5]=
2
-----
1 + n

vanishes as n tends to Infinity.

Approximations to the limit you are after can be found by determining
the coefficient of x in the Product with x replaced by 1-x, e.g.,

In[6]:= Coefficient[Product[1 - f[m][1 - x]/(m + 1), {m, 2, 10}], x]
Out[6]=
1153
----
1155

Alternatively, simple series rearrangement gives

In[7]:= 2/(1 + 10) Sum[1 - 1/Divisors[n][[2]], {n, 2, 10}] Out[7]=
1153
----
1155

Hence the number you are after is the same as

Lim[2/(1 + p) Sum[1 - 1/Divisors[n][[2]], {n, 2, p}],p->Infinity]

I am not aware of a way of computing this expression exactly.  However,
you can easily find numerical approximations:

In[8]:= Timing[N[2/(1+10000) Sum[1 - 1/Divisors[n][[2]], {n,2,10000}]]]
Out[8]= {25.9 Second, 1.33959}

In[9]:= Timing[N[2/(1+20000) Sum[1 - 1/Divisors[n][[2]], {n,2,20000}]]]
Out[9]= {65.85 Second, 1.33972}

Cheers,
Paul

____________________________________________________________________
Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia            Nedlands WA  6907
mailto:paul at physics.uwa.edu.au  AUSTRALIA
http://www.physics.uwa.edu.au/~paul

God IS a weakly left-handed dice player
____________________________________________________________________

```

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