I'm having difficulty finding an asymptotic formula for the following product:

$$ k^{\alpha}\prod_{1 \leq i \leq N \atop i \neq k} (i^{\alpha}-k^{\alpha})$$ where $N$ is a parameter tending to infinity and $\alpha = \alpha(N)$ is on the order of $N^{-1}$.

The method I've tried to use to estimate this product is by giving a second order Taylor expansion of $e^{\alpha \log i} - e^{\alpha\log k}$ with $1 \leq i, k \leq N$, and splitting the product into the ranges $1 \leq i \leq k-1$ and $k+1 \leq i \leq N$ (a higher order expansion seems unnecessary given how small $\alpha$ is and how many terms are included in the summation). Splitting the product in this fashion (after factoring out $(-1)^{k-1}\alpha^{N-1}$ and making some simpler estimates) gives rise to two products \begin{align*} \prod_{1 \leq i \leq k-1} \log(k/i) = \exp\left(\sum_{1 \leq i \leq k-1} \log_2(k/i)\right) \\ \prod_{k+1\leq i \leq N} \log(i/k) = \exp\left(\sum_{k+1 \leq i \leq N} \log_2(i/k)\right), \end{align*} (here $\log_2 t := \log(\log t)$) and it seems that Euler-Maclaurin summation is the obvious move here in order to estimate the arguments of the exponents (I hope to get, at the very least, $o(1)$ error terms for these arguments as $N \rightarrow \infty$). On the other hand, the summands in the first and second integral blow up whenever $k/i$ or $i/k$ is very close to 1 (which will occur when $N$ gets large since $k$ is meant to be generic), respectively; if we ignore terms of the form $i/k \in ((1+\delta)^{-1},1+\delta)$, where $\delta > 0$ is some parameter that we can choose then we get sums that are more or less computable via Euler-Maclaurin if $\delta$ is not chosen too small (since the integral $\int_{1+\delta}^{N/k} \log_2 t \ dt$ can be computed in relation to the logarithmic integral for example, which has well-studied asymptotic expansions).
The sum over the $\delta$-interval that we've omitted, though, is difficult to compute for general values of $k$, and it seems to me that some upper bound must be taken: for instance, i we use the trivial bound (e.g., for $k+1 \leq i \leq (1+\delta)k$ we take $|\log_2 (i/k)| \leq |\log_2(1+1/k)| \leq 2\log k$) then this middle sum gives about $\delta k \log k$ as a contribution, and clearly, then, $\delta$ must be chosen very small in order to give an error term in the exponential that gives rise to an asymptotic formula, and this is unsuitable.

I would appreciate any sort of a hint (but, please, no full solutions) to answer at least one of the following questions:

  1. is there a better approach to this estimate in order to get a small error term without exponentiation?

    1. Is there a better technique to bound the sum over the $((1+\delta)^{-1},1+\delta)$ interval than what is presented above?

Any input is sincerely appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.