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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.

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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 $$ a_n = e^{x + o(1)} \xrightarrow[n\to\infty]{} e^x\,.\tag{4} $$

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  • $\begingroup$ It seems that the asymptotic expansion of the harmonic series mentioned here math.stackexchange.com/a/596640/53441 (where $\gamma$ is the Euler–Mascheroni constant) makes the proof clear. $\endgroup$ – bkarpuz Jul 4 '18 at 23:47
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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.

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