I am trying to show that the following function has a lower bound of $\ \frac{1}{2}$ for all $c\geq 2$. Or, alternatively, that that function increases with $\ c$:

$$\frac{\Gamma\left[1+\frac{1}{c}\right] \Gamma[1+c]}{\Gamma\left[1+\frac{1}{c}+c\right]}$$

Note that this can be rewritten a few ways, such as $\ c\text{ B}[1+\frac{1}{c}, c]$ Plotting it out to 10,000 or so, it seems very clear that it is increasing, approaching 1, but I can't figure out how to prove it either through integration or properties of the special functions. Can anyone help me out? Thank you!

Note: I found in this document in Theorem 3, if $$(a-1)(b-1) \geq 0$$ then $$\text B(a,b) \geq \frac{1}{ab}$$

So, I thought that if I took the function as $\ c \text B(1+ \frac{1}{c},c)$ where $\ (\frac{1}{c})(c-1) \geq 0$ we have $$c \text B(1+ \frac{1}{c},c) \geq \frac{c}{\frac{c+1}{c} c} = \frac{c}{c+1} > \frac{1}{2}$$

However, I must be doing something wrong, as the value of $\ c \text B(1+ \frac{1}{c},c)$ when $c=2$ is actually less than $\frac{2}{3}$ (though greater than $\frac{1}{2}$)

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    $\begingroup$ The relevant part of Theorem 3 in the document you cite shows that $\Gamma(a+b)\geqslant ab\Gamma(a)\Gamma(b)$ for every $a>1$ and $b>1$. Since $\beta(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$, this is logically equivalent to $\beta(a,b)\leqslant1/(ab)$. In other words there is a typo in their (3.17). $\endgroup$ – Did Nov 4 '11 at 9:30
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    $\begingroup$ By the way, the inequality $\beta(a,b)\leqslant1/(ab)$ for $a>1$ and $b>1$ (and the similar ones in the other regimes) is a consequence of the representation of $\beta(a,b)$ as an integral over $(0,1)$ and of a simple rearrangement/coupling inequality. $\endgroup$ – Did Nov 4 '11 at 9:33

The following proof has as its key idea the use of the mean value theorem to deal with the quotient $\Gamma(1+c) / \log\Gamma(1+c+\tfrac1c)$.

Let $\psi(t) = \Gamma'(t)/\Gamma(t) = \frac d{dt}\log \Gamma(t)$ be the classical digamma function (an increasing function for $t>0$). It is known, for positive integers $n$, that $\psi(n) = H_{n-1} - \gamma$ is the difference between the $(n-1)$st harmonic number and Euler's constant; in particular, $\psi(n) < \log n$ for positive integers $n$.

It suffices to show that $-\log 2$ is a lower bound on the logarithm of your function. Note that $$ \log\Gamma(1+\tfrac1c) + \log\Gamma(1+c) - \log\Gamma(1+c+\tfrac1c) = \log\Gamma(1+\tfrac1c) - \tfrac1c \psi(t) $$ for some $t$ between $1+c$ and $1+c+\frac1c$, by the mean value theorem. $\Gamma(x)$ has a global minimum (for positive x) of about $0.8856$ near $x=1.4616$, and so $\log\Gamma(1+\tfrac1c) > -\frac18$ say. Since $\psi$ is increasing and $t\le 1+c+\frac1c < \lceil c+2 \rceil$, we have $\psi(t) < \log( \lceil c+2 \rceil ) < \log(c+3)$. Therefore $$ \log\Gamma(1+\tfrac1c) + \log\Gamma(1+c) - \log\Gamma(1+c+\tfrac1c) > -\frac18 - \frac{\log(c+3)}c, $$ which exceeds $-\log 2$ for $c\ge4$ say.

  • $\begingroup$ Thank you so much, Greg! That is incredibly helpful. One question - what does "say" mean, as in "$>-\frac{1}{8}$say" and "$\geq4$say"? $\endgroup$ – OctaviaQ Nov 4 '11 at 17:24
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    $\begingroup$ ah, it means I'm making a sort of arbitrary choice of the constant. One could do better than $-\frac18$ and $4$ if one really wanted to, but I just chose simple numbers that were pretty tight. $\endgroup$ – Greg Martin Nov 4 '11 at 19:12

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