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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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    The sum converges to the size (finite or infinite) of A. Hint: prove this for finite A; then compare the limits for A and for any subset of A.2011-08-04
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    http://en.wikipedia.org/wiki/Monotone_convergence_theorem2011-08-04
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    More interesting question: is $\lim_{x\to1^-} (1-x)\sum_{n\in A}x^n$ always the [asymptotic density](http://en.wikipedia.org/wiki/Natural_density), where such exists?2011-08-04
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    @anon: it is (more interesting and always the asymptotic density, where such exists).2011-08-04
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    You comments imply that $\lim_{x\to1^-} (1-x) \sum_{p \in \mathbb{P}} x^p = 0$, but can one find the behavior more precisely ? Is the sum diverging as a power, log, or slower than that ?2011-08-04
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    I don't know what one can say in full generality. That seems to depend strongly on the set $A$.2011-08-04
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    For $P=${primes}, $(1-x)\sum_{p\in P}x^p$ behaves as $1/|\log(1-x)|$.2011-08-04
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    @Didier Would you kindly sketch how you arrived at this result whenever you have time. Did you use the [prime number theorem](http://en.wikipedia.org/wiki/Prime_number_theorem) ? Thank you.2011-08-04
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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

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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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    Yo, wouldn't this mean $f(x)\sim A(-1/\ln x)\sim\frac{1}{(\ln x) \ln\ln(1/x)}$ as $x\to 1^{-}$?2011-08-05
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    @anon, you might know and apply to your suggested formula a simple equivalent of $\ln x$ when $x\to1^-$.2011-08-05
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    Ah, $\ln x \sim (1-x)^{-1}$, I got ya.2011-08-05
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    @anon, $\ln(x)$ does not behave like $(1-x)^{-1}$.2011-08-05
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    I mixed something up, $x-1$.2011-08-05
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    @anon, *bravo !*2011-08-05