What are the possible values of $\lim\limits_{n\to\infty} \prod_{i=1}^n (1+ y_i^{\frac 1n}/n)$? For any sequence $(y_i)_{i=1}^\infty$ of positive real numbers, let $(X_n)_{n=1}^\infty$ be another sequence defined by
$$X_n = \prod_{i=1}^n \left(1+\frac{y_i^{\frac{1}{n}}}{n}\right).$$
It is clear that $X_n >1$ for all $n$.

Question: For any $c\ge 1$, can one find a sequence $(y_i)$ of positive real numbers such that $\left(X_n\right)$ converges to $c$ as $n\to \infty$?

In an answer to a previous question, it is shown that if $y_i, y_i^{-1}$ grows slower than polynomials, then the limit of $(X_n)$ is always $e$. There are also simple examples $\left(y_i = \left(i^2\right)^i\right)$ which shows that $(X_n)$ might be unbounded.
Some remarks:

*

*One can find $(y_i)$ such that $(X_n)$ is bounded, but has no limit (this can be done by choosing $y_i$ to be $1$ most of the time, and a huge number once in a while),


*Changing finitely many members of $(y_i)$ do not alter the limit of $(X_n)$.


*Think of $(y_i)$ as a function $y: \mathbb N \to (0,\infty)$. Then $f_n : \mathbb N \to (0,\infty)$ defined by $f_n(i) = y_i^{1/n}$ is a sequence of functions which converges pointwisely to the constant function $1$. If the convergence is uniform, then limit of $(X_n)$ exists, but again the limit is $e$.
 A: Assume all functions of the form
$$
f(t)=\sum^{\infty}_{k=1}c_ke^{p_kt}.\tag 1
$$
Assume also that for all $k,n\in\textbf{N}$, we have
$$
f\left(\frac{1}{k}\right)^n=f\left(\frac{n}{k}\right).\tag 2
$$
Then set
$$
S:=\prod^{n}_{k=1}\left(1+\frac{f(k)^{1/n}}{n}\right).
$$
We have
$$
\log S=\sum^{n}_{k=1}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=-\sum^{n}_{k=1}\sum^{\infty}_{l=1}\frac{(-1)^l}{ln^l}f\left(\frac{k}{n}\right)^l=
$$
$$
=-\sum^{\infty}_{l=1}\frac{(-1)^l}{ln^{l-1}}\frac{1}{n}\sum^{n}_{k=1}f\left(\frac{k}{n}\right)^l\approx-\sum^{\infty}_{l=1}\frac{(-1)^l}{ln^{l-1}}\int^{1}_{0}f(t)^ldt=
$$
$$
=\int^{1}_{0}n\log\left(\frac{n+f(t)}{n}\right)dt.
$$
Hence when $n\rightarrow \infty$, then
$$
\lim_{n\rightarrow\infty}\prod^{n}_{k=1}\left(1+\frac{f(k)^{1/n}}{n}\right)=\exp\left(\int^{1}_{0}f(t)dt\right).\tag 3
$$
As an example we can take $f(t)=2^t$.
A: The following is a non-answer (although too long for a comment) and is in its largest part the work of Greg Martin (for their insightful comment) as well as Sangchul Lee and Carl Schildkraut for their answer to this related post.

If the limit
$$\lim_{n\to \infty}\prod_{i=1}^n\left( 1  + \frac{A^{i/n}}{n}\right)$$
can be proven to exist for any $A\ge 0$, then the answer is in the affirmative, and would easily follow from the fact that the function
$$f:(0,\infty)\to \mathbb{R}:A\mapsto \lim_{n\to \infty}\prod_{i=1}^n\left( 1 + \frac{A^{(i/n)}}{n}\right) \tag{1}$$
is monotonically increasing and has $(1, \infty)$ as range (proven at the end of the post). In particular, for any $c\ge 1$ we may find $A\in (0,\infty)$ such that the sequence $\{ A^i\}$ gives the desired result.

Regarding whether the limit in $(1)$ exists:
letting $b_i = (A^{i/n})/n$, the inequality
$$1+\sum_{i=1}^nb_i \le \prod_{i=1}^n\left( 1 + b_i\right)\le \exp \left(\sum_{i=1}^nb_i\right)\tag{2}$$
should draw our attention to the series $\sum b_i$. Observe that, letting $s_n = \sum_i^n y_i^{1/n}$, we get
\begin{split}
\lim_{n\to \infty} \sum_{i=1}^nb_i = & \lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^nA^{i/n}\\
= & \lim_{n\to \infty}\frac{s_n}{n}\\
= & \lim_{n\to \infty}\frac{s_{n+1}-s_n}{(n+1)-n} \ \ (\text{by the Stolz-Cesàro Theorem})\\
= & \lim_{n\to \infty} \left(A + \sum_{i=1}^n \left(A^{i/(n+1)}-A^{i/n}\right)\right)\\
= & \frac{A-1}{\log A} \ \ (\text{by Sangchul Lee's/Carl Schidkraut's answer)}
\end{split}
Therefore $(2)$ becomes
$$1+\frac{A-1}{\log A} \le \liminf_{n\to\infty} \prod_{i=1}^n\left( 1 + b_i\right) \le \limsup_{n\to\infty} \prod_{i=1}^n\left( 1 + b_i\right) \le \exp \left(\frac{A-1}{\log A}\right).$$
Furthermore, the equation
$$\prod_{i=1}^n \left(1+b_i\right) = \exp \left( \sum_{i=1}^n\log(1+b_i)\right)$$
implies that $\lim_{n\to\infty}\prod (1+b_i)$ exists if and only if so does $\lim_{n\to \infty} \sum \log(1+b_i)$.

The fact that the range of $f$ is $(1,\infty)$ follows from first inequality in $(2)$ together with the facts that $f(0)=1$ and that $f$ is is monotone.
A: This answer in largest is inspired by the insightful comments and answers of users at the post. I think that a small detail is missing so I will be proving it in the answer. Let $y_i = A^i$ with $A>0$,
\begin{align}
\log X_n = \sum_{k=1}^n \log \left(1 + \frac1n A^{\frac kn}\right)
\end{align}
Using the fact that for $u \in (0, 1)$, $u - \frac{u^2}{2} \le \log (1+u) \le u$, Then for $n > 1+A^2$ and $1\le k\le n$, $\frac{1}{n} A^{\frac kn} \le \frac{1+A^{2}}{n} < 1$ and
\begin{align}
\sum_{k=1}^{n} \left(\frac1n A^{\frac kn} - \frac12 \left(\frac 1n A^{\frac {k}n}\right)^2 \right)\le \log & X_n \le \sum_{k=1}^{n} \frac1n A^{\frac kn}\\
\implies \quad \frac1n \sum_{k=1}^n A^{\frac{k}{n}} - \frac{1+A^2}{2n} \le \log & X_n \le \frac1n \sum_{k=1}^n A^{\frac kn} 
\end{align}
By tending $n\to \infty$, you have that:
\begin{align}
\lim\limits_{n\to\infty} X_n &= e^{\lim\limits_{n\to\infty} \log X_n}\\
&= e^{\lim\limits_{n\to\infty} \frac1n \sum_{k=1}^n A^{\frac{k}{n}}}\\
&= e^{\int_0^1 A^x\mathrm d x} = e^{\frac{A-1}{\log A}}
\end{align}
Since $A \to \frac{A-1}{\log A}$ sends $(0, \infty)$ to $(0,\infty)$, then the result that you are looking for is immediate.
