Computing a limit involving Gammaharmonic series

Solution 1:

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}

Solution 2:

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

Solution 3:

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