Does $\lim_{n\rightarrow\infty}\sum_{k=1}^n\ln\left(\frac{k(1+\alpha)}{n-k+1+(n-k)\alpha}\right)$ exist?

Does the limit $$\lim_{n\rightarrow\infty}\sum_{k=1}^n\ln\left(\frac{k(1+\alpha)}{n-k+1+(n-k)\alpha}\right)$$ exist?

Background: I came across this limit when considering the rocket equation for a rocket with $n$ stages. Each stage has a dry mass $m_d$ and fuel mass $m_f$ and $\alpha\equiv m_f/m_d$. The limit calculates how much $\Delta v$ one gets given some $\alpha$. I was wondering if there's a limit to how much $\Delta v$ one can get out of a rocket by letting the number of stages go to infinity. For more background see this answer.

Solution 1:

Let $1+\alpha=\beta^{-1}$. We can rewrite the term within the logarithm as

$$ f(n,k,\beta)=\frac{k}{n-k+\beta} $$

We can find the finite sum

$$ S_n(\beta)=\sum\limits_{k=1}^n \ln f(n,k,\beta)=\ln\left[\prod\limits_{k=1}^n f(n,k,\beta) \right] $$

Write out a few terms of the product

$$ \prod\limits_{k=1}^n f(n,k,\beta)=\frac{1}{n+\beta-1}\cdot \frac{2}{n+\beta-2}\cdots \frac{n}{\beta} $$

Which may be written as

$$ \prod\limits_{k=1}^n f(n,k,\beta)=\frac{\Gamma(n+1)}{\Gamma(n+\beta)/\Gamma(\beta)} $$

So we have for the sum

$$ S_n(\beta)=\ln\left[\frac{\Gamma(n+1)\Gamma(\beta)}{\Gamma(n+\beta)} \right] $$

Using the asymptotic expansion of $\Gamma$ for large argument with fixed $\beta$

$$ S_n(\beta)\sim (1-\beta)\ln n \qquad , \qquad n\to \infty $$

So that the sum in logarithmically divergent unless $\beta=1$, in which case $S_n(1)=\ln\Gamma(1)=0$