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I am trying to describe the asymptotic growth of the function $$f(n) = \sum_{k = 1}^{n-1} \frac{k^{2n - 4k - 3}(n^2 -2nk + 2k^2)}{(n-k)^{2n-4k-1}}$$ as $n \rightarrow \infty$. Plotting $f(n)$ for the first 100 or so values of $n$ makes me think that $f \in \mathcal{O}(\sqrt{n})$, but I am looking for the constant $c$ such that $f(n) \sim c\sqrt{n}$, if it exists.

EDIT: Put the $\sqrt{n}$’s in the numerators of the asymptotics instead of the denominator.

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    $\begingroup$ This is quite standard, but I don't have time to explain (you can probably find hints in Knuth vol 1). The result is $f(n)\sim\sqrt{\pi n}/2$ (you probably wanted the $\sqrt{n}$ in the numerator). $\endgroup$
    – Henri Cohen
    Commented Sep 13, 2023 at 21:40
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    $\begingroup$ A discrete Laplace method en.wikipedia.org/wiki/Laplace%27s_method will work. Write the main factor $k^{2n-4k} / (n-k)^{2n-4k}$ as an exponential $e^{F(k)}$ and Taylor expand $F$ around the maximum of $k=n/2$ and this should get you the asymptotics you want after some calculations. A change of variables $k = n/2+m$ may be slightly useful (you then have to treat the case of odd and even $n$ slightly differently, but this is a minor inconvenience). $\endgroup$
    – Terry Tao
    Commented Sep 13, 2023 at 21:43
  • $\begingroup$ Thank you both @TerryTao and @HenriCohen! Though I'm a bit confused about the full application of the Laplace method. Using $F(k) = \ln(\frac{k}{n-k}^{2n - 4k})$, we get a second derivative of $F''(k) = -\frac{2n^3}{k^2(n-k)^2}$, which evaluates to $F''(\frac{n}{2}) = - \frac{2^5}{n}$. Laplace's method tells us that $\sum_{k=1}^{n-1}e^{M \cdot F(k)} \sim \sqrt{\frac{\pi n}{2^4 M}}$ as $M \rightarrow \infty$, but I don't see how that helps us estimate $f(n) \sim \sum_{k=1}^{n-1}e^{F(k)}$. $\endgroup$
    – Gabe Schoenbach
    Commented Sep 13, 2023 at 23:10
  • $\begingroup$ I recommend studying why Laplace’s method works rather than applying it as a “black box”, as this will let you adapt the method for your specific situation (as well as similar ones in the future). Don’t forget to handle the remaining factors in the summand that are not covered by $e^{F(k)}$. You may also find the Euler-Maclaurin formula useful to compare the sum with the integral. $\endgroup$
    – Terry Tao
    Commented Sep 13, 2023 at 23:47
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    $\begingroup$ One general tip with these questions is to first solve the problem by "cheating strategically", and then remove the "cheats" one by one. Examples of such cheats for your problem include: 1. Pretending that the sum and the integral are interchangeable. 2. Pretending that Laplace's method is valid even when $M=1$. 3. Pretending that Taylor expansions (to second order) of the function $F$ are exact. 4. Pretending that subexponential terms such as $n^2-2nk+2k^2$ are equal to their value at the critical point $k=n/2$. 5. Pretending that $n/2$ is always an integer. etc. $\endgroup$
    – Terry Tao
    Commented Sep 14, 2023 at 1:06

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