Product limit with exponentials

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

We can use the Euler-Maclaurin Sum Formula to get a fairly accurate value for this product.

Using Taylor series gives $$ \begin{align} &\log\left(2-e^{1/n}\right)\\ &\small=-\frac1n-\frac1{n^2}-\frac1{n^3}-\frac{13}{12n^4}-\frac5{4n^5}-\frac{541}{360n^6}-\frac{223}{120n^7}-\frac{47293}{20160n^8}-\frac{36389}{12096n^9} \end{align} $$ Applying the Euler-Maclaurin Sum Formula to this series yields $$ \begin{align} &\sum_{k=2}^n\log\left(2-e^{1/k}\right)\\ &\small\sim C-\log(n)+\frac1{2n}+\frac1{12n^2}+\frac1{36n^3}+\frac1{80n^4}+\frac1{300n^5}+\frac1{5040n^6}+\frac{17}{28224n^7}+\frac{829}{483840n^8} \end{align} $$ Thus, $$ \begin{align} &\small\log\left(n\prod_{k=2}^n\left(2-e^{1/k}\right)\right)\\ &\small\sim C+\frac1{2n}+\frac1{12n^2}+\frac1{36n^3}+\frac1{80n^4}+\frac1{300n^5}+\frac1{5040n^6}+\frac{17}{28224n^7}+\frac{829}{483840n^8} \end{align} $$ where $C$ is the log of the limit we seek and the next term in the series is approximately $\frac1{8000n^9}$.

Using $n=100$ yields $$ e^C\approx0.5335376801314199077153 $$ Using $n=1000$ yields $$ e^C\approx0.53353768013141990771530812952793 $$