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Consider the ordinary generating function of the sequence of primes ($2+3x+5x^2+7x^3 + ...$); by the ratio test and the prime number theorem, its radius of convergence is $1$. Thus, we might well ask about the limit of its value as $x$ approaches $-1$ from above (i.e., about the Abel summation of the alternating series of primes $2 - 3 + 5 - 7 + ...$). So I do! Is this limit well-defined? Is it finite? If so, what is known about its value? (I pessimistically suspect I will be told it is actually disappointingly infinite, or even worse, that the limit diverges by oscillation, but I haven't enough experience in this sort of thing to confidently rule out more interesting possibilities...)

[I realize the alternating series of primes has partial sums with arbitrarily large magnitudes, by the existence of arbitrarily large prime gaps, so there's no hope for it to be convergent in the standard sense...]

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Forget prime gaps - the terms don't go to zero, that's enough to rule out convergence in the standard sense. –  Gerry Myerson May 28 '11 at 5:40
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Ha, good point! Sometimes you get so caught up in a line of thought, you miss the most obvious routes to the same destination... –  Sridhar Ramesh May 28 '11 at 5:50
    
For fun, I asked wolfram alpha to compute the sum formed by replacing the $n$-th prime with the approximation $n \log n$. To my surprise, it gave an answer: [ \sum_{n=1}^{\infty} x^n n \log n = -\text{PolyLog}^{(1,0)}(-1, x) ] –  Hurkyl May 28 '11 at 16:01
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The point being that ${\rm PolyLog}(a,x) = \sum_{n=1}^\infty n^{-a} x^n$, and $\left. \frac{d}{da} n^{-a} \right|_{a=-1} = - n \log n$, so your sum is $\left. - \frac{d}{da} \rm{PolyLog}(a,x)\right|_{a=-1}$. –  Robert Israel May 29 '11 at 5:57
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It does appear that ${\rm PolyLog}(a,-1)$ is analytic in a neighbourhood of $a=-1$. –  Robert Israel May 29 '11 at 6:05
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2 Answers

I've a summation-method based on the triangle of Eulerian-numbers, (just for reference:
$ \qquad \qquad \tiny \begin{array} {rrrrr} 1 & . & . & . & . & . \\\ 1 & 0 & . & . & . & . \\\ 1 & 1 & 0 & . & . & . \\\ 1 & 4 & 1 & 0 & . & . \\\ 1 & 11 & 11 & 1 & 0 & . \\\ 1 & 26 & 66 & 26 & 1 & 0 \\\ ... & ... & ... \end{array} $

Call this (infinite) matrix E. The row-sums are the factorials and scaling the rows by the reciprocal factorials make the rowsums the unit; actually we can construct a regular matrix-summation method on this. We use the column-sums of E, more precisely the dot-products of the infinite vector containg our primes with alternating signs and the reciprocal factorials (call it X)

$ \qquad \qquad \small X=\text{ [ 2 , -3 , 5/2! , -7/3! , 11/4! , -13/5! , 17/6! , -19/7! , ... ]} $

with E in a resulting vector T, which contains then the sequence of the Eulerian-transforms

$ \qquad \qquad \small X * E = T $

as instance of an "Eulerian transformation" (as different to the other name of "Euler-transformation" which uses the pascalmatrix instead). Finally the sum of the entries in T is the acceptable value for the alternating sum of the primes under "Eulerian summation", if this sum converges. We want finally write, using the lower triangular matrix of ones D for the partial sums in the vector S :

$ \qquad \qquad \small D * T^{\tiny \text{ T}} = S $

and if the sequence of entries in S converges, then $s =\lim_{n\to \infty} s_n $ will be our limit for the alternating sum of primes.

Because of the reciprocal factorial which is involved in the dotproducts, because of the little growthrate of the sequence of primenumbers and because of the simple composition of the eulerian numbers each of that columnwise dotproduct occuring in the multiplication $X *E$ is convergent, so each entry in T is well defined. Here are the first couple entries of T:

$ \qquad \qquad \small T = \text{[ 0.7036728 , 1.059633 , 0.9470500 , 0.006954269 , -0.9466667 , -0.6131519 , 0.1756172, 0.5145809,... ]} $

and the partial sums from T show a nicer behave than that of the sequence of alternating signed primes. However, this is still not conclusive - maybe with some more terms (I computed this up to n=127) one can give a more conclusive result. Here is a plot of the entries in T

alt text

It impresses me, that this Eulerian transforms are all of small absolute value, and this let's me hope, that we might be on the right track here. On the other hand, in the following plot of the partial sums, the (directly summed) partial sums still show oscillating behave (its summands still having alternating signs), so I've also included three variants using additional Eulersums of small orders (0.2,1 and 2) to hopefully flatten that oscillating curves down to convergence.

alt text

However these Eulersums still does not suffice to arrive at a convergent sequence of partial sums - at least not with 128 terms only, perhaps we should look up to 256 terms or more.

[update]: After optimizing the procedures I've got the first 256 partial sums of the T-vector:

partial sums S n=256

(Well, from the pictures I begin to suspect that we need still another method, possibly fourier-series or even a completely different approach.)


[update]: to answer on Greg's comment I compared the profile of partial sums of that summation using the Eulerian numbers and that of the common, ordinary Euler-summation. Surprise: the profile is nearly identical, only that the method using Eulerian summation arrives at the same profile using only the forth part of coefficients:

comparision

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Doing this "Eulerian summation" is basically looking at a weighted average of an interval of primes that doesn't start at 1. For example, if the nth row of the Eulerian triangle has the property that most of its mass is concentrated within about $C\sqrt n$ of $n/2$, then Eulerian summation is basically computing the alternating sum of the primes from $n/2-C\sqrt n$ to $n/2+C\sqrt n$. This is not going to yield information about the original posted problem. –  Greg Martin May 29 '11 at 16:34
    
@Greg: hmm, I don't really understand the critique, I'll have to think about this again. Anyway, this summation method proved useful in summing the geometric series in the halfplane $R<1$ and even allows to assign the correct value to the alternating series of factorials - something that Euler-summation cannot accomplish. So I don't understand yet why my answer seems to be a useless. (Even your comment sounds a bit disgusted - so my answer may have been obfuscating? - so sorry for that) –  Gottfried Helms May 29 '11 at 19:45
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Numerical evidence is somewhat bleak:

  • $x = -0.9$ yields $2.18185604596612$.
  • $x = -0.99$ yields $10.8600238817757$.
  • $x = -0.999$ yields $-42.1972332872236$.
  • $x = -0.9999$ yields $-197.508471042688$.
  • $x = -0.99999$ yields $-2299.82561828947$.
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