Computing a limit involving Gammaharmonic series 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! 
 A: 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:
$$\gamma+\sum_{n=1}^{+\infty}\left(\frac{1}{\Gamma(1/n)}-\frac{1}{n}\right)=0.8188638872713\ldots\tag{3}$$
A: $(1)\enspace$ A simplification of what Leucippus had written, for a better understanding:
$\displaystyle \lim_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{\Gamma(1/k)} - \ln\Gamma\left(\frac{1}{n}\right)\right) =\lim_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{\Gamma(1/k)} - \ln n \right) =$
$\displaystyle =\gamma + \sum\limits_{k=1}^\infty \left(\frac{1}{\Gamma(1/k)} - \frac{1}{k}\right) = \gamma + \sum\limits_{n=2}^\infty a_n \zeta(n)\enspace$ for $\displaystyle\enspace \frac{1}{\Gamma(z)}-z =: \sum\limits_{n=2}^\infty a_n z^n$
$(2)\enspace$ Let $\,\displaystyle \tau:=-\int_0^1\left\lfloor \frac{1}{t} \right\rfloor\left(1+\frac{\psi(t)}{\Gamma(t)}\right) dt \,$ 
with the Digamma function $\,\displaystyle\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}\,$ and Gauss's floor function $\,\lfloor x \rfloor\,$. 
Then we get with $\,\Gamma(1/n)\thicksim n\,$ : 
$$\lim_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{\Gamma(1/k)} - \ln\Gamma\left(\frac{1}{n}\right)\right)=\lim_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{k} - \ln\Gamma\left(\frac{1}{n}\right)\right) +\sum\limits_{k=1}^\infty \left(\frac{1}{\Gamma(1/k)} - \frac{1}{k}\right)$$
$$=\gamma+\tau \hspace{2.1cm}$$  
A: One can show that
Lemma 1. 
If $f(x)$ is analytic in $(-a,a)$, $a\geq 1$, then
$$
\sum^{M}_{n=1}f\left(\frac{1}{n}\right)=\int^{M}_{1}f\left(\frac{1}{t}\right)dt+c(f)+O\left(\frac{1}{M}\right)\textrm{, }M\rightarrow +\infty,\tag 1
$$
where $c_f$ is a constant depended from $f$ and not from $M$:
$$
c_f=f(0)+f'(0)\gamma+\sum^{\infty}_{k=2}\frac{f^{(k)}(0)}{k!}\left(\zeta(k)-\frac{1}{k-1}\right).
$$
Proof.
Expand $f$ in (1) into Taylor series, then sum and integrate. The ''infinite'' terms involving $M$ are canceled and the result will follow.  
Remark. The next two known estimates are usefull in the proof of Lemma 1:
$$
\sum^{M}_{k=1}\frac{1}{k}=\log(M)+\gamma+O\left(\frac{1}{M}\right)\textrm{, }M\rightarrow \infty
$$ 
and
$$
\zeta_n(s)-\zeta(s)=O\left(\frac{1}{n^{s-1}}\right)\textrm{, }s>1\textrm{, }n\rightarrow\infty.
$$
Lemma 2.
If $f(x)=xg(x)$, $g(0)=1$ and $g$ analytic in $(-a,a)$, $a\geq 1$, then
$$
\int^{M}_{1}f\left(\frac{1}{t}\right)dt+\log\left(f\left(\frac{1}{M}\right)\right)=c'_f+o(1)\textrm{, }M\rightarrow +\infty,
$$
where
$$
c'_f=\sum^{\infty}_{k=1}\frac{f^{(k+1)}(0)}{k(k+1)(k)!}
$$
Proof.
We can write
$$
\int^{M}_{1}f\left(\frac{1}{t}\right)dt=\int^{1}_{1/M}\frac{f(t)}{t^2}dt=\int^{1}_{h}\frac{g(t)}{t}dt=
$$
$$
=-g(0)\log(h)+\sum^{\infty}_{k=1}\frac{g^{(k)}(0)}{k(k)!}(1-h^k).
$$
Hence
$$
\lim_{M\rightarrow \infty}\left[\int^{M}_{1}f\left(\frac{1}{t}\right)dt+\log\left(f\left(\frac{1}{M}\right)\right)\right]
=
$$
$$
=\lim_{h\rightarrow 0}\left[\int^{1}_{h}\frac{g(t)}{t}dt+\log\left(hg(h)\right)\right]=
$$
$$
=\lim_{h\rightarrow 0}\left[-g(0)\log(h)+\log(hg(h))\right]+\sum^{\infty}_{k=1}\frac{g^{(k)}(0)}{k(k)!}=\sum^{\infty}_{k=1}\frac{f^{(k+1)}(0)}{k(k+1)(k)!}
$$
QED
From the two Lemma's we have
Theorem.
If $f(x)=xg(x)$, $g(0)=1$ and $g$ analytic in $(-a,a)$, $a\geq 1$, then 
$$
\sum^{M}_{k=1}f\left(\frac{1}{k}\right)+\log\left(f\left(\frac{1}{M}\right)\right)=c''_f+o(1),
$$
with constant term
$$
c''_f=f'(0)\gamma+\sum^{\infty}_{k=2}\frac{f^{(k)}(0)}{k!}\zeta(k)
$$
Note: The estimate $o(1)$ can easily replaced with $O\left(\frac{1}{M}\right)$.
Hence as one can see the problem can be generalized. In your case we have:
The function $f(z)=\frac{1}{\Gamma(z)}$ is entire and have Taylor expansion in (-1,1). Set
$$
g(x)=\frac{f(x)}{x}=\frac{1}{\Gamma(x+1)}.
$$ 
Then $f(0)=0$, $g(0)=1$, $f'(0)=1$ and hence holds
$$
\sum^{M}_{n=1}\frac{1}{\Gamma\left(\frac{1}{n}\right)}+\log\left(\frac{1}{\Gamma\left(\frac{1}{M}\right)}\right)=\gamma+\sum^{\infty}_{k=2}a_k\zeta(k),
$$
or equivalent
$$
\sum^{M}_{n=1}\frac{1}{\Gamma\left(\frac{1}{n}\right)}-\log\left(\Gamma\left(\frac{1}{M}\right)\right)=\gamma+\sum^{\infty}_{k=2}a_k\zeta(k)+O\left(\frac{1}{M}\right)\textrm{, }M\rightarrow\infty,
$$
where 
$$
a_k=\frac{1}{k!}\left(\frac{d^k}{dx^k}\frac{1}{\Gamma(x)}\right)_{x=0}\textrm{, }k=2,3,\ldots
$$
A: 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}
