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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...

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2 Answers 2

up vote 8 down vote accepted

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, $$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq 1- \frac{c}{c+1} = \frac{1}{c+1}.$$

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 $$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = 1 - \frac{c}{c}\left(1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)\right)\left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)$$ $$= \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$

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If you don't like integration by parts, you can just use $\int_0^1 t (1-t)^a \mathrm{d}t = \int_0^1 (1-t)^a \mathrm{d}t - \int_0^1 (1-t)^{a+1} \mathrm{d}t$. –  cardinal Oct 24 '11 at 12:44
I added the integration by parts calculation to show that it's not difficult, as there seems to be some implication that it is somehow to be avoided. –  Mike Spivey Oct 24 '11 at 13:40
Funny enough, I only decided to add the comment based on your original wording which I read as your apparent displeasure with having to resort to integration by parts (maybe because of the parenthetical remark that followed). So, I was just trying to show a simple way to avoid it. Obviously, I appear to have misinterpreted that original statement. Cheers. :) –  cardinal Oct 24 '11 at 13:57
@MikeSpivey I apologize if my answer created the impression that I feel that integration by parts is to be avoided for any reason. But, making the connection with the Gamma function and using the recursion does have the merit that we get a formula that anyone who can do arithmetic can use to calculate the exact value of $Beta(c+1,1/c)$, and of course, it also yields the bound desired by the OP as well. Of course, as I said upfront in my answer, to get to the result one does need to know something about Beta and Gamma functions that may not be known to, say, first-semester calculus students. –  Dilip Sarwate Oct 24 '11 at 13:59
@cardinal: And it sounds like I was reading something into your comment that wasn't there! :) I can see how my original parenthetical remark made it look like I was unhappy with using integration by parts. I wasn't unhappy, though; I was just giving another reference. Thanks for your comment, too; that's a nice way to rewrite the integral. (The current upvote on the comment is from me.) –  Mike Spivey Oct 24 '11 at 16:45

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, $$\text{Beta}(c+1, 1/c) < \frac{c^2}{c+1} \Rightarrow \frac{1}{c}\text{Beta}(c+1, 1/c) < \frac{c}{c+1} = 1 - \frac{1}{c+1}$$

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Thanks for your comment on my answer, and I agree that it's helpful to see the inequality from the point of view of the gamma function representation. (In fact, it appears that both of our answers essentially reduce to showing - in different ways - that $\text{Beta}(c+1,1/c) < \text{Beta}(2,1/c) = c^2/(c+1)$. The current upvote on your answer is from me, by the way.) –  Mike Spivey Oct 24 '11 at 16:51

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