Show that $$\lim_{n\rightarrow+\infty}\sum_{k=1}^n \displaystyle \left( \Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}$$ where $\gamma$ is the Euler-Mascheroni Constant.

Motivation : One can show that $$\lim_{n\rightarrow+\infty}\displaystyle\left(n-\Gamma\bigl(\frac{1}{n}\bigr)\right)=\gamma.$$ This means that $\Gamma\bigl(\frac{1}{n}\bigr)\sim n$ when $n$ is large. So we have that (even if is not correct) $\Gamma\bigl(\frac{k}{n}\bigr)\sim \frac{n}{k}$. It implies that $$\sum_{k=1}^{n}\displaystyle\left(\Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}\sim \sum_{k=1}^{n}(\frac{k}{n})^k.$$ Since the limit of the right sum exists and its value is $\frac{e}{e-1}$. Numerical calculations show that the limit of the sum involving the Gamma function would be $\frac{e^\gamma}{e^\gamma− 1}$.

  • $\begingroup$ Just a small thought: the sum looks (aside from the exponential term) suspiciously like a Riemann sum, and I wonder whether there's an approach that tackles it from that direction... $\endgroup$ – Steven Stadnicki Aug 17 '14 at 19:08
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:00

Step 1. If $$I_1(n)=\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_1(n)=0$.

Proof. Indeed, since $\Gamma$ is decreasing on $(0,1]$ we have $$ I_1(n)\leq\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}\leq\sum_{k=1}^\infty\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}=\frac{1}{\Gamma(1/\sqrt{n})-1}$$ and step 1. follows.

Step 2. If $$I_2(n)=\sum_{\sqrt{n}<k\leq n/2}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_2(n)=0$.

Proof. Recall that $\Gamma$ attains its minimum $\approx0.8856$, on $[1,2]$, at some some point $x_0\approx1.4616$. In particular, $\Gamma(x)\geq2/3$ for $1\leq x\leq 2$. So, for $\sqrt{n}<k\leq n/2$ we have $$ \frac{k}{n}\Gamma\left(\frac{k}{n}\right)=\Gamma\left(1+\frac{k}{n}\right) \geq\frac{2}{3} $$ Thus, for $\sqrt{n}<k\leq n/2$, we have $\Gamma(k/n)>4/3$. It follows that $$ I_2(n)\leq \sum_{k>\sqrt{n}}\left(\frac{3}{4}\right)^k=4\left(\frac{3}{4}\right)^{\lceil\sqrt{n}\rceil} $$ and step 2. follows.

Step 3. If $$I_3(n)=\sum_{n/2<k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_3(n)=\dfrac{e^\gamma}{e^\gamma-1}$. where $\gamma$ is the Euler-Mascheroni constant.

Proof. Note first that, with $p=n-k$, $$ I_3(n)=\sum_{0\leq p<n/2}\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n} =\sum_{p=0}^\infty a_p(n) $$ with $$a_p(n)=\left\{\matrix{\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n}&\hbox{if}& 0\leq p<n/2\cr0&\hbox{otherwise}}\right.$$ Now, since $\Gamma(1)=1$ and $\Gamma'(1)=-\gamma$ we have, for a fixed $p$ and large $n$: $$(p-n)\ln\Gamma\left(1-\frac{p}{n}\right)=(p-n)\ln\left(1+\frac{\gamma p}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)=-\gamma p+\mathcal{O}\left(\frac{1}{n}\right)$$ Thus $$ \forall\,p\geq 0,\quad \lim_{n\to\infty}a_p(n)=e^{-\gamma p}.\tag{1} $$ Now, we will need the next lemma.

Lemma. For $t\in[1/2,1]$ we have $(\Gamma(t))^{t/(1-t)}\geq \Gamma(1/2)=\sqrt{\pi}.$

Taking, this lemma for granted, we conclude by taking $t=1-p/n$ when $0\leq p<n/2$, that $$ \forall\,p\geq 0,n\geq 1,\quad a_p(n)\leq \left(\frac{1}{\sqrt{\pi}}\right)^p. \tag{2} $$ and clearly, $$\sum_{p=0}^\infty \left(\frac{1}{\sqrt{\pi}}\right)^p<+\infty\tag{3}$$ Combining $(1)$, $(2)$ and $(3)$ we conclude that $$ \lim_{n\to\infty}I_3(n)=\lim_{n\to\infty}\sum_{p=0}^\infty a_p(n) =\sum_{p=0}^\infty\lim_{n\to\infty}a_p(n)= \sum_{p=0}^\infty e^{-\gamma p}=\frac{e^\gamma}{e^\gamma-1}.$$

The desired conclusion follows: $$ \lim_{n\to\infty}\sum_{1\leq k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}= \lim_{n\to\infty}(I_1(n)+I_2(n)+I_3(n))=\frac{e^\gamma}{e^\gamma-1}. $$

Proof of the Lemma. Let $f(t)=\dfrac{t}{1-t}\ln\Gamma(t)$. Then $f'(t)=\dfrac{g(t)}{(1-t)^2}$ with $$g(t)=\ln\Gamma(t)+t(1-t)\psi(t);\quad\hbox{where $\psi(t)=\Gamma'(t)/\Gamma(t)$}$$ and $g'(t)=(1-t)h(t)$ with $$h(t)=2\psi(t)+t\psi'(t)$$ and finally $h'(t)=3\psi'(t)+t\psi''(t)=\sum_{k=0}^\infty\frac{3k+t}{(k+t)^3}>0$.

So, $h$ is increasing, and $\lim_{t\to0^+}h(t)=-\infty$, $h(1)=\frac{\pi^2}{6}-2\gamma>0$. This proves that $h(t)<0$ for $0<t<x_0$ and $h(t)>0$ for $x_0<t<1$, for some $x_0$.

And $g$ is decreasing on $[0,x_0]$ and increasing on $[x_0,1]$. But $\lim_{t\to0^+}g(t)=+\infty$, $g(1)=0$. This proves that $g$ has exactly one change of sign on $(0,1)$ from positive to negative. This proves that the minimum of $f$ on $[1/2,1]$ is $\min(f(1/2),f(1))=f(1/2)$, and the lemma is proved.

  • 3
    $\begingroup$ +1, I hope you enjoyed going into such detail. I wouldn't. $\endgroup$ – Start wearing purple Aug 18 '14 at 10:17
  • $\begingroup$ @O.L. Even if you wouldn't, your (now deleted ?) answer doesn't give any ideas how can we prove the equivalent. $\endgroup$ – Krokop Aug 18 '14 at 10:27
  • $\begingroup$ @Krokop No, it doesn't. And? $\endgroup$ – Start wearing purple Aug 18 '14 at 10:44
  • $\begingroup$ @O.L. And is not useful. $\endgroup$ – Krokop Aug 18 '14 at 10:54
  • 1
    $\begingroup$ Ovidiu Furdui gives this problem as an open problem in his book "Limits, series and fractional part integrals". $\endgroup$ – aziiri Aug 18 '14 at 18:03

Since $\log(\Gamma(x))$ is convex, $\log(\Gamma(x))\ge-\gamma(x-1)$ and $\log(\Gamma(x))\ge(1-\gamma)(x-2)$. Thus, $\log(\Gamma(x))\ge-\gamma+\gamma^2$. That is, $\Gamma(x)\ge e^{-\gamma+\gamma^2}=0.78345806514\gt\frac34$.

For $1\le k\le\frac23n$, $$ \begin{align} \Gamma\left(\frac kn\right)^{-k} &=\left(\frac kn\right)^k\,\Gamma\left(1+\frac kn\right)^{-k}\\ &\le\left(\frac{4k}{3n}\right)^k\tag1 \end{align} $$ Since $\left(1+\frac1k\right)^k\lt e\lt\left(1+\frac1k\right)^{k+1}$, we have $$ \frac4{3n}ke\le\frac4{3n}\frac{(k+1)^{k+1}}{k^k}\le\frac4{3n}(k+1)e\tag2 $$ Therefore, $\left(\frac{4k}{3n}\right)^k$ decreases while $\frac kn\lt\frac{3}{4e}$, then it increases.

Thus, applying $(1)$ and $(2)$, $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^{2n/3}\Gamma\left(\frac kn\right)^{-k} &\le\lim_{n\to\infty}\overbrace{\ \ \ \ \frac4{3n}\ \ \ \ \vphantom{\left(\frac89\right)^{\!\frac23n}}}^{k=1}+\overbrace{\frac23n\max\!\left(\frac{64}{9n^2},\left(\frac89\right)^{\!\frac23n}\right)}^{2\le k\le\frac23n}\\ &=0\tag3 \end{align} $$

For $0\le k\le\frac13n$, since $\Gamma\left(\frac{n-k}n\right)\ge1+\frac{\gamma k}n$ and $\left(1+\frac{\gamma k}n\right)^{n+\gamma k}\ge e^{\gamma k}$, $$ \begin{align} \Gamma\left(\frac{n-k}n\right)^{k-n} &\le\left(1+\frac{\gamma k}n\right)^{k-n}\\[3pt] &\le e^{-\gamma k\frac{n-k}{n+\gamma k}}\\[9pt] &\le e^{-\frac{2\gamma}{3+\gamma}k}\tag4 \end{align} $$ Furthermore, since $\Gamma\left(\frac{n-k}n\right)=1+\frac{\gamma k}n+O\left(\frac kn\right)^2$, $$ \lim_{n\to\infty}\Gamma\left(\frac{n-k}n\right)^{k-n}=e^{-\gamma k}\tag5 $$ by Dominated Convergence, $(4)$ and $(5)$ show that $$ \begin{align} \lim_{n\to\infty}\sum_{k=2n/3}^n\Gamma\left(\frac kn\right)^{-k} &=\lim_{n\to\infty}\sum_{k=0}^{n/3}\Gamma\left(\frac{n-k}n\right)^{k-n}\\ &=\sum_{k=0}^\infty e^{-\gamma k}\\[3pt] &=\frac{e^\gamma}{e^\gamma-1}\tag6 \end{align} $$

Putting $(3)$ and $(6)$ together, we get $$ \lim_{n\to\infty}\sum_{k=1}^n\Gamma\left(\frac kn\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}\tag7 $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy