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}

1$\begingroup$ Where does this arise? What makes you think this is true? $\endgroup$– Yemon ChoiFeb 18, 2013 at 19:36

$\begingroup$ A professor gave us this exercise, but my colleagues and I weren't able to solve it, even if we found it very interesting. $\endgroup$– FeliceFeb 18, 2013 at 22:13

3$\begingroup$ So why not ask this professor? $\endgroup$– Yemon ChoiFeb 19, 2013 at 1:31
1 Answer
$$ g(\epsilon)=\sum_{k\ge 1} k^se^{\frac {k1}2\ln(1\epsilon)} \lesssim \int_0^{+\infty} x^s e^{a\epsilon x} dx= \int_0^{+\infty}x^s e^{ax}dx\epsilon^{s1} $$ where $a$ is a fixed constant. So $$ C=\int_0^{+\infty}x^s e^{ax}dx,\quad \phi(s)=s1. $$

$\begingroup$ 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 $\epsilon$. $\endgroup$– FeliceFeb 18, 2013 at 21:52

$\begingroup$ Anyway I have some difficulty understanding your inequalities. For example: \begin{equation} e^{\frac{x1}{2}ln(1\epsilon)}\leq e^{a\epsilon x}. \end{equation} Is that inequality true for $x=\frac{1}{2}$ and $\epsilon \rightarrow 0$? $\endgroup$– FeliceFeb 18, 2013 at 22:04


$\begingroup$ The only problem is when $\epsilon$ is small, and you do have a singularity at $\epsilon=0$. Now for $x>1,0<\epsilon<1/4$, $$ \frac{x1}{2}\ln(1\epsilon)\le\frac{x(\epsilon/2)}{2} $$ so you can take $a=1/4$. The bounded values of $x$ are unimportant. $\endgroup$– BazinFeb 19, 2013 at 8:25