Calculation of the value of $ \lim_{ n→∞}a^{-nk} ∏_{j=1}^k(a +j/n)^n$ 
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.
 A: 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,
$$S_n = a^{-nk} \prod_{j=1}^k \left(a + \dfrac{j}n \right)^n = \prod_{j=1}^k \left(1 + \dfrac{j}{an} \right)^n$$
$$\lim_{n \to \infty} S_n = \underbrace{\prod_{j=1}^k \lim_{n \to \infty} \left(1 + \dfrac{j}{an} \right)^n = \prod_{j=1}^k \exp(j/a)}_{\text{By using the fact that }\lim_{n \to \infty} \left(1 + \frac{x}n\right)^n = e^x} = \exp\left(\dfrac{\sum_{j=1}^k j}{a} \right) = \exp\left(\dfrac{k(k+1)}{2a} \right)$$
A: $$
\begin{align}
\log\left(a^{-nk}\prod_{j=1}^k(a+j/n)^n\right)
&=n\left(-k\log(a)+\sum_{j=1}^k\log(a+j/n)\right)\\
&=n\sum_{j=1}^k\log\left(1+\frac{j}{an}\right)\\
&=n\sum_{j=1}^k\left(\frac{j}{an}+O\left(\frac1{n^2}\right)\right)\\
&=\frac{k(k+1)}{2a}+O\left(\frac1n\right)
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}a^{-nk}\prod_{j=1}^k(a+j/n)^n=\exp\left(\frac{k(k+1)}{2a}\right)
$$
A: 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 $n \to \infty$.
