Limit of a sequence defined by a product Let $x>0$, and set
\begin{equation}
a_{n}:=\prod_{j=\lfloor\frac{n}{\mathrm{e}}\rfloor}^{n}\left(1+\frac{x}{j}\right),\quad{}n\geq3,
\end{equation}
where $\lfloor\cdot\rfloor$ is the floor function.
Question. Show that $\lim_{n\to\infty}a_{n}=\mathrm{e}^{x}$.
Clearly, the sequence is not monotonic.
Thus, what comes to my mind is the squeezing theorem.
I can show that $a_{n}\leq\bigl(\frac{n}{\lfloor\frac{n}{\mathrm{e}}\rfloor-1}\bigr)^{\lambda}$ for $n\geq6$, and $\bigl(\frac{n}{\lfloor\frac{n}{\mathrm{e}}\rfloor-1}\bigr)^{x}\to\mathrm{e}^{x}$ as $n\to\infty$.
Although it seems to be correct, I cannot show $a_{n}>\mathrm{e}^{x}$ for all large $n$.
I would be glad if you can help in this direction.
 A: Rewrite
$$
a_n = \exp\left( \sum_{j=\lfloor n/e\rfloor}^n \ln\left(1+\frac{x}{j}\right) \right)\tag{1}
$$
Since $x$ is fixed, we have $\lim_{n\to \infty}\frac{x}{j} = 0$ for every $\left\lfloor \frac{n}{e}\right\rfloor\leq j\leq n$, and thus can write
$$
 \ln\left(1+\frac{x}{j}\right) = \frac{x}{j} + O\left(\frac{1}{j^2}\right)
$$
using the Taylor expansion of $\log(1+u)$ when $u\to0$. We get
$$
\sum_{j=\lfloor n/e\rfloor}^n \ln\left(1+\frac{x}{j}\right)
= x\sum_{j=\lfloor n/e\rfloor}^n \frac{1}{j} + \sum_{j=\lfloor n/e\rfloor}^n O\left(\frac{1}{j^2}\right)
= x\left( \log n - \log \left\lfloor \frac{n}{e}\right\rfloor+o(1)\right) + o(1) \tag{2}
$$
the last equality using the asymptotic expansion of the Harmonic series twice:
$$
H_n = \log n + \gamma + o(1)
$$
as well as the fact that the serie $\sum_j \frac{1}{j^2}$ converges, implying that the "Cauchy slice" $\sum_{j=\lfloor n/e\rfloor}^n \frac{1}{j^2}$ converges to $0$ as $n\to\infty$.
From (2), it is not hard, e.g. by the Squeeze theorem to handle the floor, to show that
$$
\sum_{j=\lfloor n/e\rfloor}^n \ln\left(1+\frac{x}{j}\right)
= x\sum_{j=\lfloor n/e\rfloor}^n \frac{1}{j} + \sum_{j=\lfloor n/e\rfloor}^n O\left(\frac{1}{j^2}\right)
= x \log e + o(1) = x + o(1) \tag{3}
$$
so that, plugging this in (1), we finally obtain
$$
a_n = e^{x + o(1)} \xrightarrow[n\to\infty]{} e^x\,.\tag{4}
$$
A: It is simpler to take a look at the sequence
$$b_n = \log a_n = \sum_{j=\lfloor n/e\rfloor} ^n \log\left(1+ \frac{x}{j} \right)\;.$$
We would like to show that $b_n \to x$ for $n\to \infty$. 
The function $\log (1+x/j)$ is monotonously decreasing as a function of $j$. As a result, we can find the bounds
$$ \tag{1}\int_{\lfloor n/e\rfloor -1}^n \log\left(1+ \frac{x}{y} \right) dy\leq b_n \leq \int_{\lfloor n/e\rfloor}^{n+1} \log\left(1+ \frac{x}{y} \right) dy \;. $$
We can evaluate the integral
$$\int_{\alpha n + O(1)}^{\beta n + O(1)} \log\left(1+ \frac{x}{y}\right) dy = n \int_{\alpha+O(1/n)}^{\beta+O(1/n)} \log\left(1+\frac{x}{nz}\right)dz \stackrel{(n\to\infty)} \to x\log(\beta/\alpha)\;.$$
In (1), we have $\beta=1$ and $\alpha =1/e$ both in the lower as well as in the upper bound. The result
$$b_n \to x\log(\beta/\alpha) =x$$
follows.
