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}
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$$ g(\epsilon)=\sum_{k\ge 1} k^se^{\frac {k1}2\ln(1\epsilon)} \lesssim \int_0^{+\infty} x^s e^{a\epsilon x} dx= \int_0^{+\infty}x^s e^{ax}dx\epsilon^{s1} $$ where $a$ is a fixed constant. So $$ C=\int_0^{+\infty}x^s e^{ax}dx,\quad \phi(s)=s1. $$ 

