Beta Function -- finding a lower bound based on parameters

Question

I would like to show that $$ 1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq \frac{1}{c+1}.$$ for all $c \geq 2$. I have plotted it out for $c$ up through 200, and it seems to hold. Does anyone have any tips on if this can be formally shown? I imagine an expert on the beta function might know...

Answer 1

The definition of the beta function is $$Beta\left(c+1,\frac{1}{c}\right) = \int_0^1 t^c (1-t)^{1/c-1} dt.$$ Since $c \geq 2$ and $t \in [0,1]$, $t^c \leq t$. Thus (applying integration by parts to the first integral), we have $$Beta\left(c+1,\frac{1}{c}\right) \leq \int_0^1 t (1-t)^{1/c-1} dt = \left. -c t (1-t)^{1/c}\right|_0^1 + c\int_0^1 (1-t)^{1/c} dt$$ $$= \left. \frac{-c (1-t)^{1/c+1}}{1+1/c} \right|_0^1 = \frac{c}{1+1/c} = \frac{c^2}{c+1}.$$ Therefore, $$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq 1- \frac{c}{c+1} = \frac{1}{c+1}.$$

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The bound isn't very tight as $c$ increases, though, as we can see from the asymptotics: $$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$

Derivation: $$Beta\left(c+1,\frac{1}{c}\right) = \frac{\Gamma(c+1) \Gamma \left(\frac{1}{c}\right)}{\Gamma\left(c+1+\frac{1}{c}\right)} = \frac{c \Gamma(c+1) \Gamma \left(1+\frac{1}{c}\right)}{\Gamma\left(c+1+\frac{1}{c}\right)}.$$ Now, $\Gamma \left(1+\frac{1}{c}\right) = 1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)$. (This is via the Maclaurin series for $\Gamma(x+1)$; see Expression 8.321 in Gradshteyn and Ryzhik.)

Also (see the DLMF), $$\frac{\Gamma(c+1)}{\Gamma\left(c+1+\frac{1}{c}\right)} = c^{-1/c} \left(1 + O\left(\frac{1}{c^2}\right)\right) = \exp\left(\frac{-\log c}{c}\right)\left(1 + O\left(\frac{1}{c^2}\right)\right)$$ $$= \left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)\left(1 + O\left(\frac{1}{c^2}\right)\right) = \left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right).$$

Putting this all together, we have $$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = 1 - \frac{c}{c}\left(1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)\right)\left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)$$ $$= \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$

Answer 2

I was composing the following when the original answer by @MikeSpivey appeared; his edited answer provides considerably more information. The short calculation below arrives at the desired result without requiring integration by parts but does need knowledge of the relationship between the Beta and Gamma functions.

For $c \geq 2$ $$\begin{align*} \text{Beta}(c+1, 1/c) &= \frac{\Gamma(c+1)\Gamma(1/c)}{\Gamma(c + 1 + 1/c)}\\ &= \frac{c}{c+1/c}\times \frac{c-1}{c-1+1/c}\times \cdots \times \frac{2}{2+1/c}\times \frac{1}{1+1/c} \times \frac{1}{(1/c)\Gamma(1/c)}\times \Gamma(1/c). \end{align*}$$ The product of the last three terms on the right is $c^2/(c+1)$ while all the other terms have value less than $1$. Hence, $$\text{Beta}(c+1, 1/c) < \frac{c^2}{c+1} \Rightarrow \frac{1}{c}\text{Beta}(c+1, 1/c) < \frac{c}{c+1} = 1 - \frac{1}{c+1}$$