Product limit with exponentials Find an explicit formula for the limit: 
$$\lim_{n \rightarrow \infty} n \prod_{k=2}^{n} (2 - e ^ {\frac 1 k})$$ 
I am not asking for convergence proof since I know the sequence is decreasing and bounded.
 A: This is going to be a very primitive attempt to answer this question. Yet I think it is worth posting since this kind of calculations appear many times in calculus. We take the log of  the limit and then we Taylor expand the following function:
\begin{equation}
\log\left(2 - e^{\frac{1}{k}}\right) = \log\left(1 - \sum\limits_{p=1}^\infty \frac{1}{p!} \frac{1}{k^p}\right) = -\frac{1}{1} \sum\limits_{p=1}^\infty \frac{1}{p!} \frac{1}{k^p} - \frac{1}{2} \sum\limits_{p=2}^\infty \frac{2^p-2}{p!} \frac{1}{k^p}-\frac{1}{3} \sum\limits_{p=3}^\infty \frac{3^p-3 \cdot 2^p+3}{p!} \frac{1}{k^p}-\frac{1}{4} \sum\limits_{p=4}^\infty \frac{4^p-4 \cdot 3^p+6\cdot 2^p-4}{p!} \frac{1}{k^p}-\cdots
\end{equation}
Now if we denote the unknown limit by $g$ we clearly have:
\begin{equation}
\log(g) = \log(n) + \sum\limits_{k=2}^n \log(2-e^{\frac{1}{k}}) = \log(n) - (H_n-1) - \frac{1}{1} \sum\limits_{p=2}^\infty \frac{1}{p!} (\zeta(p)-1)) - \frac{1}{2} \sum\limits_{p=2}^\infty \frac{2^p-2}{p!} (\zeta(p)-1) -\frac{1}{3} \sum\limits_{p=3}^\infty \frac{3^p-3 \cdot 2^p+3}{p!} (\zeta(p)-1) -\cdots
\end{equation}
Now $\log(n) - (H_n-1) \rightarrow 1 - \gamma$ when $n\rightarrow \infty$ and the remaining sums all converge . The final result is therefore:
\begin{equation}
g = \exp\left(1-\gamma - \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{1}{p!} \sum\limits_{l=1}^p \frac{1}{l} \sum\limits_{j=0}^{l-1} \binom{l}{j} (l-j)^p (-1)^j \right) = \exp\left(1-\gamma - \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{1}{p!} \sum\limits_{l=1}^p \frac{1}{l} l! S_2(p,l)\right) = \exp\left(1-\gamma - \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{1}{p!} (-1)^p Li_{1-p}(2)\right) \simeq 0.5335376801314199077153...
\end{equation}
where $S_2(p,l)$ are Stirling numbers of the second kind and $Li_n(x)$ is the polylogarithmic function.
The question remains is it possible to further simplify the result..
Note that the original sequence converges very slowly. It is only for $n> 60000000$ that $g_n < 0.53353768$. On the other hand the series in the exponential in the last formula converge very fast. I truncated the sum over $p$ at $p=150$ and I obtain an accuracy of twenty three digits. 
Now, the question remains is it possible to further simplify the result. Let us take the sum in the exponential:
\begin{equation}
S = \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{(-1)^p}{p!} Li_{1-p}(2) = \sum\limits_{k=2}^\infty \sum\limits_{p=2}^\infty \frac{(-1/k)^p}{p!} Li_{1-p}(2) = \sum\limits_{k=2}^\infty \sum\limits_{p=2}^\infty \frac{(-1/k)^p}{p!}  \frac{d^{p-1}}{d \epsilon^{p-1}} \left.\left(\frac{2 e^{\epsilon}}{1-2 e^{\epsilon}}\right)\right|_{\epsilon=0} = \sum\limits_{k=2}^\infty \int\limits_{0}^{-1/k} \left(\frac{2 e^\xi}{1-2 e^\xi} - \frac{2}{1-2}\right) d\xi = \sum\limits_{k=2}^\infty \left( \log\frac{1}{2 e^{-1/k}-1} - \frac{2}{k}\right) = \sum\limits_{k=2}^\infty \left(-\log(2-e^{1/k})-\frac{1}{k}\right) = - \log(g) + \log(n) - \left(H_n-1\right)
\end{equation}
Taking the limit $n\rightarrow \infty$ we get $S = -\log(g) + 1- \gamma$ which is equivalent to $g = \exp(1-\gamma-S)$. As we can see this was going in circles. But at least, as a sanity check, we made sure that the result is correct.
A: A few hints that might help you out: If the product in your expression is $P = e^{\log P}$, then you can use the property of the logarithm $\log (2-e^{\frac{1}{k}}) = \log 2 +\log (1-\frac{e^{\frac{1}{k}}}{2})$. For large $k$ the last expression is close to $-\log 2$, so you can use Taylor series for expansion. 
