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For all $s>0$ define for $\epsilon\in(0,1)$ the function: \begin{equation} g(\epsilon)=\sum_{k=0}^{\infty}(1+k)^s(\sqrt{1-\epsilon})^k. \end{equation} Prove that $\exists C>0$ and $\phi(s)$ such that: \begin{equation} g(\epsilon)\leq C \epsilon^{\phi(s)}. \end{equation}

flag	asked Feb 18 at 18:54 Felice 25●1●3

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Where does this arise? What makes you think this is true? – Yemon Choi Feb 18 at 19:36

A professor gave us this exercise, but my colleagues and I weren't able to solve it, even if we found it very interesting. – Felice Feb 18 at 22:13

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So why not ask this professor? – Yemon Choi Feb 19 at 1:31

Bazin · Accepted Answer · 2013-02-19 08:20:37Z UTC

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$$ g(\epsilon)=\sum_{k\ge 1} k^se^{\frac {k-1}2\ln(1-\epsilon)} \lesssim \int_0^{+\infty} x^s e^{-a\epsilon x} dx= \int_0^{+\infty}x^s e^{-ax}dx\epsilon^{-s-1} $$ where $a$ is a fixed constant. So $$ C=\int_0^{+\infty}x^s e^{-ax}dx,\quad \phi(s)=-s-1. $$

edited Feb 19 at 8:20

answered Feb 18 at 21:02

Bazin
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	I think that the constant $a$ depends on $\epsilon$. I would like to find an estimate such that the constant $C$ doesn't depend on $\epsilon$. – Felice Feb 18 at 21:52
	Anyway I have some difficulty understanding your inequalities. For example: \begin{equation} e^{\frac{x-1}{2}ln(1-\epsilon)}\leq e^{-a\epsilon x}. \end{equation} Is that inequality true for $x=\frac{1}{2}$ and $\epsilon \rightarrow 0$? – Felice Feb 18 at 22:04
	$\epsilon \rightarrow 1$, sorry. – Felice Feb 18 at 22:06
	The only problem is when $\epsilon$ is small, and you do have a singularity at $\epsilon=0$. Now for $x>1,0<\epsilon<1/4$, $$ \frac{x-1}{2}\ln(1-\epsilon)\le\frac{x(-\epsilon/2)}{2} $$ so you can take $a=1/4$. The bounded values of $x$ are unimportant. – Bazin Feb 19 at 8:25

An interpolation inequality.

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