Convergence of $\sum_{n=1}^{\infty} a^{-\frac{1}{c+1}}\prod_{k=1}^{n} \frac{a}{a+k^{c}}$ as $a\to\infty$ It looks to me, by doing numerical simulations, that
$$
f(c) = \lim_{a\rightarrow \infty}a^{-\frac{1}{c+1}}\sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{a}{a+k^{c}}
$$
converges and that $1\leq f(c) <2$, which was quite surprizing to me. It is easy to see that $f(0)=1$ (using geometric series) and it has been shown that $f(1)= \sqrt{\frac{\pi}{2}}$, as shown by the answer to Convergence of $ \sum_{k=1}^{\infty} \sqrt{a} \prod_{i=1}^{k}\frac{1}{1+i a}$ as $a \rightarrow 0$. I also understand that $\lim_{c\rightarrow \infty} f(c) = 1$ as only the first term of the sum is non-zero. I tried to prove the convergence for all $c\geq0$ using standard convergence tests, or by looking at $f'(c)$, but without success.
Also, I wonder if it's possible to find a "nicer" (i.e. without the limit) expression for $f(c)$, but that's probably too much to ask.
Any help would be much appreciated, thanks!
ps. If helpful, I can post the results of the simulations
 A: I am going to prove the following claim:

Claim. For $c > 0$ and $\alpha = 1/(c+1)$, we have
  $$ f(c) = \alpha^{1-\alpha} \Gamma(\alpha). $$

Proof. We may focus on the case $c > 0$ and we assume so. Let us plug $a = u^{c+1}$. Then we get
$$ S = S(c,a) := a^{-1/(c+1)} \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{a}{a+k^{c}}
= \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{1}{1+(k^{c}/u^{c+1})} \frac{1}{u}. $$
The main idea of the proof is the observation that the last infinite series above behaves like a Riemann sum as $u \to \infty$. To this end, fix $M > 0$ and divide the sum into two parts:
$$ S = \sum_{n \leq Mu} \prod_{k=1}^{n} \frac{1}{1+(k^{c}/u^{c+1})} \frac{1}{u} + \sum_{n > Mu} \prod_{k=1}^{n} \frac{1}{1+(k^{c}/u^{c+1})} \frac{1}{u} =: S_{1} + S_{2}. $$
First, let $N = \lfloor Mu \rfloor$. If $n > Mu$, then $n > N$ and
\begin{align*}
\prod_{k=1}^{n}\frac{1}{1+(k^{c}/u^{c+1})}
&\leq \prod_{k = N+1}^{n} \frac{1}{1+(k^{c}/u^{c+1})} \\
&\leq \prod_{k = N+1}^{n} \frac{1}{1+M^{c}/u}
= \left( \frac{1}{1+M^{c}/u} \right)^{n-N}.
\end{align*}
This shows that
$$ S_{2}
\leq \frac{1}{u} \sum_{n=1}^{\infty} \left( \frac{1}{1+M^{c}/u} \right)^{n} = \frac{1}{M^{c}}. $$
Next, if $n \leq N$ then 
\begin{align*}
\log \prod_{k=1}^{n} \frac{1}{1+(k^{c}/u^{c+1})} 
&= -\sum_{k=1}^{n} \log \left(1+\frac{k^{c}}{u^{c+1}}\right) \\
&= -\int_{0}^{n} \log \left(1+\frac{t^{c}}{u^{c+1}}\right) \, dt + \mathcal{O}\left(\tfrac{M^{c}}{u} \right) \\
&= -\int_{0}^{n/u} u \log \left(1+\frac{t^{c}}{u}\right) \, dt + \mathcal{O}\left(\tfrac{M^{c}}{u} \right)  \\
&= -\frac{(n/u)^{c+1}}{c+1} + \mathcal{O}\left(\tfrac{M^{2c+1}}{u} \right).
\end{align*}
Plugging this back,
$$ S_{1} = \sum_{n \leq Mu} \left(1 + \mathcal{O}\left(\tfrac{M^{2c+1}}{u} \right)\right) \exp\left( -\frac{(n/u)^{c+1}}{c+1} \right) \frac{1}{u}. $$
Taking $u \to \infty$, it follows that
$$ \lim_{u\to\infty} S_{1} = \int_{0}^{M} e^{-x^{c+1}/(c+1)} \, dx. $$
Combining two observations, we obtain
$$ \int_{0}^{M} e^{-x^{c+1}/(c+1)} \, dx - \frac{1}{M^{c}}
\leq \liminf_{u\to\infty} S
\leq \limsup_{u\to\infty} S
\leq \int_{0}^{M} e^{-x^{c+1}/(c+1)} \, dx + \frac{1}{M^{c}}. $$
Taking $M \to \infty$ we finally obtain that $f(c) = \lim S$ exists and is equal to
$$ f(c) = \int_{0}^{\infty} e^{-x^{c+1}/(c+1)} \, dx = \alpha^{1-\alpha} \Gamma(\alpha). $$
