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The sum of natural numbers $ \sum_{n>0} x^n = \frac{x}{1-x}$, so as $x\to1^-$ it diverges as $(1-x)^{-1}$. So I wondered what would happen if we make the summation set thinner, i.e. $\sum_{n \in A} x^n$ for $A \subset \mathbb{N}$.


EDIT After receiving helpful suggestions, I guess I understand it a little better. If $A$ is a finite subset, than $ \lim_{x\to 1^-} \sum_{n \in A} x^n = \vert A \vert$. Thus

$ \lim_{x\to 1^-} \frac{\sum_{k \in (A \cap (1,n))} x^k}{\sum_{k=1}^n x^k} = \frac{\vert A \vert}{n} $

Taking the limit $n\to \infty$ the ratio gives the asymptotic density of $A$ in $\mathbb{N}$ if it exists. Indeed

$ \sum_{n=1}^\infty x^{2n} = \frac{x^2}{1-x^2} \;\;\; \text{ therefore } \;\;\; \lim_{x->1^-} (1-x) \frac{x^2}{1-x^2} = \frac{1}{2} $

What I still do not understand is how would I find the asymptotic behavior of the sum in case when the limit $\frac{\vert A\vert}{n}$ goes to zero, like in the case of $A$ being the set of prime numbers $A = \mathbb{P}$.

I guess the technique to be used here is to apply Mellin transform, and related the divergence rate to properties of zeta function.

$ \sum_{n>=1} \int_0^\infty t^{s-1} e^{-t n} dt = \Gamma(s) \sum_{n>=1} n^{-s} = \Gamma(s) \zeta(s) $

In the case of prime set this give $\Gamma(s) \zeta_\mathbb{P}(s)$ which goes as $log(s-1)$ for $s\to 1^+$, but I do not see it through yet..

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    @anon By Littlewood's Tauberian theorem, the converse is also true. If $\lim_{x\to1^-} (1-x)\sum_{n\in A}x^n$ exists, then so does the asymptotic density, and the two coincide.2011-08-05

1 Answers 1

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One is interested in the behaviour when $x\to1$ of the function $f(x)=\sum\limits_na_nx^n$, where $a_n=[n\,\text{prime}]$. The function $A$ defined on $t\ge0$ by $A(t)=\sum\limits_na_n[n\le t]$ enumerates the prime numbers up to $t$ hence one knows that $A(t)\sim t/\log(t)$ when $t\to+\infty$.

Now, $f(\text{e}^{-s})$ is the value at $s$ of the Laplace transform of the nonnegative measure with infinite mass $\text{d}A$. A general result on the behaviour of Laplace transforms (which, I believe, is explained in the book The Laplace transform by David V. Widder) asserts that, when $s\to0$, $ f(\text{e}^{-s})\sim A(1/s). $ In our case, when $x\to1$, $ (1-x)f(x)\sim (1-x)A(1/(1-x))\sim1/|\log(1-x)|. $

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    @anon, *bravo !*2011-08-05