# An interpolation inequality.

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

• Where does this arise? What makes you think this is true? Feb 18, 2013 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. Feb 18, 2013 at 22:13
• So why not ask this professor? Feb 19, 2013 at 1:31

$$g(\epsilon)=\sum_{k\ge 1} k^se^{\frac {k-1}2\ln(1-\epsilon)} \lesssim \int_0^{+\infty} x^s e^{-a\epsilon x} dx= \int_0^{+\infty}x^s e^{-ax}dx\epsilon^{-s-1}$$ where $a$ is a fixed constant. So $$C=\int_0^{+\infty}x^s e^{-ax}dx,\quad \phi(s)=-s-1.$$
• 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 $\epsilon$. Feb 18, 2013 at 21:52
• Anyway I have some difficulty understanding your inequalities. For example: $$e^{\frac{x-1}{2}ln(1-\epsilon)}\leq e^{-a\epsilon x}.$$ Is that inequality true for $x=\frac{1}{2}$ and $\epsilon \rightarrow 0$? Feb 18, 2013 at 22:04
• $\epsilon \rightarrow 1$, sorry. Feb 18, 2013 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. Feb 19, 2013 at 8:25