Computing a limit involving Gammaharmonic series

Question

It's a well-known fact that $$\lim_{n\to\infty} (H_n-\log(n))=\gamma.$$

If I use that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder if it's possible to compute the following limit in a closed-form

$$\lim_{n\to \infty}\left(\frac{1}{ \Gamma\left(\displaystyle \frac{1}{1}\right)}+ \frac{1}{ \Gamma\left( \displaystyle \frac{1}{2}\right)}+ \cdots + \frac{ 1}{ \Gamma \left( \displaystyle \frac{1}{ n}\right) }- \log\left( \Gamma\left(\displaystyle\frac{1}{n}\right)\right)\right),$$ where I called $\displaystyle \sum_{k=1}^{\infty}\frac{ 1}{ \Gamma \left( \displaystyle \frac{1}{ k}\right) }$ as Gammaharmonic series.
I can get approximations, but I cannot get the precise limit, and I don't even know if it can be expressed in terms of known constants.

A 500 points bounty moment: I would enjoy pretty much finding a solution (containing a closed-form) for the posed limit, hence the generous bounty. It's unanswered for 3 years and 8 months, and it definitely deserves another chance. Good luck!

Answer 1

From the Weierstrass product for the Gamma function we have, as $x\to+\infty$: $$\frac{1}{\Gamma(1/x)}=\frac{1}{x}+\frac{\gamma}{x^2}+O\left(\frac{1}{x^3}\right)\tag{1}$$ and: $$\log\Gamma(1/x)=\log x -\frac{\gamma}{x}+O\left(\frac{1}{x^2}\right)\tag{2}$$ gives that the value of the limit is: $$\gamma+\sum_{n=1}^{+\infty}\left(\frac{1}{\Gamma(1/n)}-\frac{1}{n}\right)=0.8188638872713\ldots\tag{3}$$

Answer 2

From Wolfram Gamma Function equations (35)-(37) provide \begin{align}\tag{1} \frac{1}{\Gamma(x)} = x + \gamma x^{2} + \sum_{k=3}^{\infty} a_{k} x^{k} \end{align} where, $a_{1}=1$, $a_{2}=\gamma$,
\begin{align}\tag{2} a_{n} = n a_{1} a_{n-1} - a_{2} a_{n-2} + \sum_{k=2}^{n} (-1)^{k} \zeta(k) \, a_{n-k}. \end{align} Now, \begin{align}\tag{3} \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} \approx H_{n} + \gamma H_{n,2} + \sum_{k=3}^{\infty} a_{k} H_{n,k}, \end{align} where $H_{n,r}$ are the generalized Harmonic numbers given by \begin{align}\tag{4} H_{n,r} = \sum_{s=1}^{n} \frac{1}{s^{r}}. \end{align} Since the limit is for large values of $n$, $n \rightarrow \infty$, then utilize the approximation, Wolfram Harmonic Number Approximations, \begin{align}\tag{5} H_{n,r} \approx \frac{(-1)^{r} \psi^{(r-1)}(1)}{(r-1)!} - \frac{1}{(r-1) \, n^{r-1} } \left( 1 + \mathcal{O}\left(\frac{1}{n}\right) \right) \end{align} to obtain \begin{align}\tag{6} \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} \approx H_{n} - \frac{\gamma}{n} + \sum_{k=2}^{\infty} \frac{(-1)^{k} a_{k}}{(k-1)!} \, \psi^{(k-1)}(1) + \mathcal{O}\left(\frac{1}{n^{2}} \right). \end{align}

Since, \begin{align}\tag{7} - \ln \Gamma\left( \frac{1}{n} \right) \approx \frac{\gamma}{n} - \ln(n) + \mathcal{O}\left(\frac{1}{n^{2}}\right) \end{align} then \begin{align}\tag{8} \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} - \ln \Gamma\left( \frac{1}{n} \right) \approx H_{n} - \ln(n) + \sum_{k=2}^{\infty} \frac{(-1)^{k} a_{k}}{(k-1)!} \, \psi^{(k-1)}(1) + \mathcal{O}\left(\frac{1}{n^{2}} \right). \end{align} Taking the limit as $n \rightarrow \infty$ and using \begin{align} \lim_{n \rightarrow \infty} \left( H_{n} - \ln(n) \right) = \gamma \end{align} then \begin{align}\tag{9} \lim_{n \rightarrow \infty} \left[ \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} - \ln \Gamma\left( \frac{1}{n} \right) \right] = \sum_{k=1}^{\infty} \frac{(-1)^{k} a_{k}}{(k-1)!} \, \psi^{(k-1)}(1). \end{align} Since \begin{align}\tag{10} \psi^{(m)}(x) = (-1)^{m+1} m! \zeta(m+1, x) \end{align} then \begin{align}\tag{11} \lim_{n \rightarrow \infty} \left[ \sum_{r=1}^{n} \frac{1}{\Gamma\left(\frac{1}{r}\right)} - \ln \Gamma\left( \frac{1}{n} \right) \right] = \gamma + \sum_{k=2}^{\infty} a_{k} \zeta(k). \end{align}