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Re: Re: Programming an Infinite Sum


Dr Bob wrote:
> 
> This matches all but the first term of your series --- I think!  (You
> didn't have things in Mathematica notation, so I found it a little
> confusing.)
> 
> ClearAll[term, sum]
> term[3] = 2/5;
> term[n_Integer] /; n > 3 := term[n] = term[n - 1](Prime[n - 1] - 2)
> /Prime[n]
> term[n_] := term@Round@n (* for plots *)
> sum[3] = term[3];
> sum[n_Integer] /; n > 3 := sum[n] = sum[n - 1] + N@term[n]
> sum[n_] := sum@Round@n  (* for plots *)
> term /@ Range[3, 8]
> Short[sum /@ Range[3, 5002], 5]
> 
> {2/5, 6/35, 6/77, 54/1001, 54/1547, 810/29393}
> 
> {0.4, 0.571429, 0.649351, 0.703297, \[LeftSkeleton]4992\[RightSkeleton],
>    0.978568, 0.978569, 0.97857, 0.97857}
> 
> (Add 1/5 to get the result with your first term.)
> 
> The ratio test is inconclusive:
> 
> Plot[(Prime[Round[n]] - 2)/
>     Prime[Round[n] + 1], {n, 4, 1000}];
> 
> a constant times
> 1/(n*Log[n]^2.7),
> 
> Plot[n*Log[n]^2.7*term[n], {n, 3, 1000}];
> 
> but I don't have a ready proof of it, despite the plot.
> 
> Likewise, it seems that THAT series converges.
> 
> Sum[1/(n*Log[n]^2.7), {n, 2, 100000}]
> 
> 1.98995
> 
> It certainly does if the following is correct:
> 
> NSum[1/(n*Log[n]^2), {n, 2, Infinity}]
> 
> 2.109742801134059 + 0.*I
> 
> but I wouldn't bet a gold mine on it, without a theoretical proof.
> 
> Bobby
> 
> On Sat, 23 Aug 2003 08:09:04 -0400 (EDT), A.S. Haley <ashaley at nccn.net>
> wrote:
> 
> > I have a tricky infinite sum over the primes and want to see whether it
> > converges or diverges.  I am having difficulty programming it (in
> > Mathematica 4), and would appreciate some help.  The sum is as follows:
> >
> > 3/5 + (3/5)(2/7) + (3/5)(5/7)(2/11) + (3/5)(5/7)(9/11)(2/13) + . . .
> >
> > + [(Prime[3]-2)(Prime[4]-2). . .(Prime[n-1]-2)/(Prime[3])(Prime[4]) . .
> > .(Prime[n-1])] x [2/(Prime[n])]  . . .
> >
> > and I want to see what happens as n --> Infinity.
> >
> > Thanks in advance,
> >
> > Allan Haley
> >
> >
> >
> 
> --
> majort at cox-internet.com
> Bobby R. Treat

Here is a nonrigorous approach that, I believe, could be made rigorous
with appropriate majorization. Among other things it gives an asymptotic
estimate for terms that is close to what you found numerically. I'll use
'k' to denote generic constants that may change from one step to
another.

The idea is to approximate Prime[j] by j*Log[j]. A rigorous argument
would need to find appropriate upper and lower bounds to this
approximation, among other things.

The nth term, for large n, may be approximated as

k * Product[1-2/(j*Log[j]),{j,m,n}] * 2/(n*Log[n])

where m is some suitably large fixed integer (and k is dependent on m).

The logarithm of this term is thus described as

k + Sum[Log[1-2/(j*Log[j]),{j,m,n}] - (Log[n] + Log[Log[n]])

This is approximated as

k - Sum[2/(j*Log[j]),{j,m,n}] - (Log[n] + Log[Log[n]])

which in turn is approximated as

k - 2*Log[Log[n]] - (Log[n] + Log[Log[n]])

Hence the nth term is approximated by the exponential of this, or

k * 1/n * 1/(Log[n])^3

This converges by the integral test.


Daniel Lichtblau
Wolfram Research


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