Let $n$ be a positive integer, and let $c$ be a positive constant. Consider the asymptotic estimate $$ \sum_{1\le kv < n}\frac{ve^{c\sqrt{n-kv}}}{n-kv} = e^{c\sqrt{n}}\left(1 + \mathcal{O}\!\left(\frac{1}{n^{\frac{1}{2} + \varepsilon}}\right)\right) $$ for some fixed $\varepsilon > 0$ and $k,v \in \mathbb{Z}^+$ are index variables and the sum is taken over all $k,v$ satisfying $1\le kv<n$.
Apparently, this estimate can be proved by Taylor expanding the denominator of the summand, extending the sum to infinity and then estimating the exponential sums that show up. But this is very tedious. I am looking for a proof of this estimate that is as short and elegant as possible. Improvements to the estimate, if possible, or any other observations and remarks are also welcome. Thank you in advance!
I think below is a correct version of asymptotic equality even if a derivation is not all rigorous. Due to the conditions $n \gg 1$ and $c > 0$, main input to the sum under consideration is from relatively small $k$ and $v$. Hence, we'll use the following approximate equalities $$ \sqrt{n-kv} \approx \sqrt{n} - \frac{kv}{2\sqrt{n}},\quad \frac1{n-kv}\approx \frac1n. $$ We need two terms of expansion for square root because of the fast rate of change of the exponential function. After these approximations, the summation can be extended to infinity. Thus, we get $$ \sum_{1\leq kv < n} \frac{v e^{c\sqrt{n-kv}}}{n - vk} \approx \frac{e^{c\sqrt{n}}}{n} \sum_{v=1}^\infty v\left(\sum_{k=1}^\infty e^{-ckv/2\sqrt{n}}\right).\quad (1) $$ For the inner sum, we have exact equality $$ \sum_{k=1}^\infty e^{-ckv/2\sqrt{n}} = \frac{e^{-cv/2\sqrt{n}}}{1 - e^{-cv/2\sqrt{n}}}. \quad (2) $$ Now (1) and (2) give $$ \sum_{1\leq kv < n} \frac{v e^{c\sqrt{n-kv}}}{n - vk} \approx \frac{e^{c\sqrt{n}}}{n} \sum_{v=1}^\infty \frac{ve^{-cv/2\sqrt{n}}}{1 - e^{-cv/2\sqrt{n}}} \approx \frac{e^{c\sqrt{n}}}{n} \int\limits_0^\infty \frac{ve^{-cv/2\sqrt{n}}}{1 - e^{-cv/2\sqrt{n}}} dv $$ Integral is related to $\zeta(2)$ and the final approximate equality is $$ \sum_{1\leq kv < n} \frac{v e^{c\sqrt{n-kv}}}{n - vk} \approx e^{c\sqrt{n}} \frac{4}{c^2} \frac{\pi^2}{6}. $$ It is not hard to numerically compute the exact sum and it's an approximate value. I see that the larger are $c$ and $n$ the better is the approximate formula. For example, for $c = 1$, $n = 1000$ values are $$ \sum_{1\leq kv < n} \frac{v e^{c\sqrt{n-kv}}}{n - vk} = 3.67 \cdot 10^{14}, \quad e^{c\sqrt{n}} \frac{4}{c^2} \frac{\pi^2}{6} = 3.56\cdot 10^{14}. $$ Relative error in this case is $0.03$. And for $c = 2$, $n = 10000$ values are $$ \sum_{1\leq kv < n} \frac{v e^{c\sqrt{n-kv}}}{n - vk} = 1.1910 \cdot 10^{87}, \quad e^{c\sqrt{n}} \frac{4}{c^2} \frac{\pi^2}{6} = 1.886\cdot 10^{87}. $$ The relative error, in this case, is $0.002$. So, probably the correct version of asymptotic equality is $$ \sum_{1\leq kv < n} \frac{v e^{c\sqrt{n-kv}}}{n - vk} = e^{c\sqrt{n}} \frac{2\pi^2}{3c^2}\left(1 + o(1)\right). \quad(3) $$
I think that somebody more mathematically educated can easily do a rigorous proof of (3) and get better asymptotics.