# A limit of a sum related to integer lattice and power series

I have the following lemma that I would like to find a source to cite for. Let $L$ be a subset of $\mathbb Z^d_{>0}$. I would like to claim that the limit $$\lim_{z \to (1,\ldots,1)^-} (\sum_{v \in L} z^v) \prod_{i=1}^d(1-z_i)$$ exists. Here $z^v$ mean $z_1^{v_1}z_2^{v_2}\ldots z_d^{v_d}$, with $v=(v_1,\ldots, v_d)$ a vector in $\mathbb Z^d$.

For the 1-dimensional case (d=1), an argument using Cesaro sum (http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation) will show that the limit exists, and I believe it can be extended to higher dimension as well, though it can be very messy, and I would like to avoid that. Does anyone know a cleaner method (or even better, a source to cite directly from) to solve this problem? Thanks in advance.

• What does $z^v$ mean, when $z$ and $v$ are vectors? – Gerry Myerson Apr 24 '14 at 5:33
• Here, $v$ is a vector and $z$ is a polynomial in multivariable. So $z^v$ mean $z_1^{v_1}z_2^{v_2}\ldots z_d^{v_d}$, with $v=(v_1,\ldots, v_d)$ a vector in $\mathbb Z^d$. I have changed the post to add this. Thanks for the comment. – sweehong Apr 24 '14 at 5:44
• Are you sure that the limit exists? Even in case $d=1$, I have doubts. Your limit seems to be related to the asymptotic density of the subset $L$. If for example $L$ is the union of the sets $\{2\cdot 2^n,\dots,3\cdot 2^n-1\}$ ($n\geq 0$), then $L$ does not have a well defined density and some numerical computations I did suggest that your limit does not exist. – Andreas Maurischat Apr 24 '14 at 14:18
• Thanks for your reply. I agree with your comment, it may not even be clear for $d=1$, and I find a flaw in my argument using Cesaro sum. It will be interesting to see if the counterexample you gave is indeed a counterexample (I believe it is). – sweehong Apr 25 '14 at 16:04

If the asymptotic density of $L\subset \mathbb{N}$ exists, then the considered limit exists and equals this density. This can be seen as follows: For $n\in\mathbb{N}$, let $L_{\leq n}:=L\cap \{1,\dots, n\}$. Then $$D:=\lim_{n\to \infty} \frac{\#L_{\leq n}}{n}$$ is the asymptotic density of $L$ (assuming it exists).
If $\frac{a}{b}\in\mathbb{Q}$ is a rational number bigger than $D$, then for $n\gg 0$, $\#L_{\leq n}< \frac{an}{b}$. In particular for $n=kb$, there are less than $ak$ numbers in $L$ smaller or equal to $kb$. Therefore, fix a $k\in\mathbb{N}$ big enough, then for all $l\geq k$ and $r\in \{0,\dots, a\}$, the $(al+r)$-th number in $L$, call it $n_{al+r}$, is bigger than $bl+r$.
Hence for all $0<z<1$: $$\sum_{l=k}^\infty \sum_{r=0}^{a-1} z^{n_{al+r}} \leq \sum_{l=k}^\infty \sum_{r=0}^{a-1} z^{bl+r} = \left(\sum_{l=k}^\infty z^{bl}\right)\left(\sum_{r=0}^{a-1} z^r \right) =\frac{z^{bk}}{1-z^b}\frac{1-z^a}{1-z}$$ and so $$(1-z)\sum_{n\in L}z^n\leq (1-z)\sum_{j<ak}z^{n_{ak}}+ z^{bk}\frac{1-z^a}{1-z^b}$$ As the last sum tends to $\frac{a}{b}$ as $z$ tends to $1$, one gets $$\limsup_{z\to 1^-} \ (1-z)\sum_{n\in L}z^n\leq \frac{a}{b}$$ for all $\frac{a}{b}>D$. Similarly, for any rational number $\frac{c}{d}<D$, one gets the estimate $$\liminf_{z\to 1^-} \ (1-z)\sum_{n\in L}z^n\geq \frac{c}{d}$$ Hence, the limit exists and is equal to $D$.
• I think we also have the converse to be true for $d=1$ i.e. If the asymptotic density of $L$ does not exist, then $\lim_{z\to 1^-} \sum_{v \in L} z^v (1-z)$ does not exist. I am thinking that by using Borel summation (en.wikipedia.org/wiki/Borel_summation), the sum $\sum_{v \in L} z^v (1-z)= \sum_{k=1}^{\infty} a_kz^k$ is equivalent to $\int_0^{\infty} e^{-t} BA(tz) dt$ for $|z|<1$, with $BA(t)=\sum_{k=1}^{\infty} \frac{a_k}{k!}t^k$. The integral is a continuous function of $z$, and if we choose our series to be nasty enough, we can probably show that the integral diverge for $z=1$. – sweehong Apr 25 '14 at 16:46