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!

• This is a very interesting question, indeed ! Numerically, it seems that there is a limit but how to express it, that is the question ! – Claude Leibovici Sep 27 '14 at 16:53
• Wrench (1968) has a paper concerning the series $\sum\frac{1}{\Gamma(k)}$. Some answers are already showing some promise to computing that series. – Ali Caglayan Sep 27 '14 at 22:56
• Are you sure that your question can be answered during 7 days although it's unanswered for 3 years and 8 months ? I think the answers of Leucippus and Jack D'Aurizio are well done! :-) – user90369 May 29 '18 at 11:56
• @user90369 no problem then. In general, one user will remain with my legacy, that is the bounty, and I'm sure they (those receiving such bounties) will want to continue to work on my question, maybe for months, years (if the case, of course), until they finally get the solution (at least out of curiosity). :-) – user 1357113 May 29 '18 at 13:04

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}

• So $a_k$ is the generating function of $\frac1{\Gamma(x)}$? – Ali Caglayan Sep 27 '14 at 22:42
• @Alizter $a_{k}$ are the coefficients in the "generating function" for $(\Gamma(x))^{-1}$. – Leucippus Sep 28 '14 at 3:51
• (+1) for the efforts so far – user 1357113 May 28 '18 at 11:42

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: $$\gamma+\sum_{n=1}^{+\infty}\left(\frac{1}{\Gamma(1/n)}-\frac{1}{n}\right)=0.8188638872713\ldots\tag{3}$$

• (+1) for the efforts so far – user 1357113 May 28 '18 at 11:41

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 $$a_k=\frac{1}{k!}\left(\frac{d^k}{dx^k}\frac{1}{\Gamma(x)}\right)_{x=0}\textrm{, }k=2,3,\ldots$$

$(1)\enspace$ 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 $\,\Gamma(1/n)\thicksim n\,$ :

$$\lim_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{\Gamma(1/k)} - \ln\Gamma\left(\frac{1}{n}\right)\right)=\lim_{n\to\infty} \left(\sum\limits_{k=1}^n \frac{1}{k} - \ln\Gamma\left(\frac{1}{n}\right)\right) +\sum\limits_{k=1}^\infty \left(\frac{1}{\Gamma(1/k)} - \frac{1}{k}\right)$$

$$=\gamma+\tau \hspace{2.1cm}$$