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Let $a > 0$ and $k ∈ \mathbb{N}$. Evaluate:
$$ \lim_{ n→∞}a^{-nk} ∏_{j=1}^k(a +j/n)^n$$

How can I solve this problem? I am totally helpless.

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Note that $$\prod_{j=1}^k \left(a + \dfrac{j}n \right)^n = a^{nk} \prod_{j=1}^k \left(1 + \dfrac{j}{an} \right)^n$$ Hence, $$S_n = a^{-nk} \prod_{j=1}^k \left(a + \dfrac{j}n \right)^n = \prod_{j=1}^k \left(1 + \dfrac{j}{an} \right)^n$$ $$\lim_{n \to \infty} S_n = \underbrace{\prod_{j=1}^k \lim_{n \to \infty} \left(1 + \dfrac{j}{an} \right)^n = \prod_{j=1}^k \exp(j/a)}_{\text{By using the fact that }\lim_{n \to \infty} \left(1 + \frac{x}n\right)^n = e^x} = \exp\left(\dfrac{\sum_{j=1}^k j}{a} \right) = \exp\left(\dfrac{k(k+1)}{2a} \right)$$

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$$ \begin{align} \log\left(a^{-nk}\prod_{j=1}^k(a+j/n)^n\right) &=n\left(-k\log(a)+\sum_{j=1}^k\log(a+j/n)\right)\\ &=n\sum_{j=1}^k\log\left(1+\frac{j}{an}\right)\\ &=n\sum_{j=1}^k\left(\frac{j}{an}+O\left(\frac1{n^2}\right)\right)\\ &=\frac{k(k+1)}{2a}+O\left(\frac1n\right) \end{align} $$ Therefore, $$ \lim_{n\to\infty}a^{-nk}\prod_{j=1}^k(a+j/n)^n=\exp\left(\frac{k(k+1)}{2a}\right) $$

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The basic idea in robjohn's proof is to use the expansion for $\ln(1+x)$ and truncate it.

The series is $\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1}x^n/n $ for $0 < x < 1$. Since the series is alternating, absolutely convergent, with the terms decreasing, it is bounded by taking any two consecutive sums. In particular, $x > \ln(1+x) > x - x^2/2$. This allows us to get explicit bounds for the sum.

Putting this into robjohn's sum, $\sum_{j=1}^k\ln\left(1+\frac{j}{an}\right) < \sum_{j=1}^k \frac{j}{an} = \frac{1}{a n}\frac{k(k+1)}{2} $ and $\sum_{j=1}^k\ln\left(1+\frac{j}{an}\right) > \sum_{j=1}^k \left( \frac{j}{an} - \frac{j^2}{2a^2 n^2}\right) = \frac{1}{a n}\frac{k(k+1)}{2} - \frac{1}{2a^2 n^2}\frac{k(k+1)(2k+1)}{6} $ so $\log\left(a^{-nk}\prod_{j=1}^k(a+j/n)^n\right) < \frac{1}{a }\frac{k(k+1)}{2} $ and $\log\left(a^{-nk}\prod_{j=1}^k(a+j/n)^n\right) > \frac{1}{a }\frac{k(k+1)}{2} - \frac{1}{2a^2 n}\frac{k(k+1)(2k+1)}{6} $.

This shows the desired result by letting $n \to \infty$.

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