0
$\begingroup$

So, right now I am writing my master thesis and I need to find a reference for a formula I found in a paper: $$ \sum_{k=1}^{\infty}k^{-\alpha}(1-\varepsilon)^k\sim b+c\Gamma(1-\alpha)\varepsilon^{\alpha-1}\text{ as }\varepsilon\downarrow0\text{ for }\alpha\in(1,2) $$ where $b,c$ are some constants. Unfortunately I have not been able to find a reference for this result, does somebody have an idea where I can find this?

Furthermore I would like to extend this result in the following sense: Let $(a_k)_{k\in\mathbb{N}_0}$ be a probability sequence, i.e. $\sum_{k\in\mathbb{N}_0}a_k=1$ such that $a_k=k^{-\alpha+o(1)}$ as $k\to\infty$. Is it possible to show that then $$ \sum_{k=1}^{\infty}a_k(1-\varepsilon)^k\sim 1+c\Gamma(1-\alpha)\varepsilon^{\alpha-1}\text{ as }\varepsilon\downarrow 0? $$ As you may have guessed by now, the subject of my thesis is probability theory and I dont know much about polylogarithm functions and its asymptotic behavior, so I thought it would be a good idea to ask here and hope that someone could help me out with this problem. Thank you for your help!

$\endgroup$

1 Answer 1

2
$\begingroup$

See for example this reference: $$\sum_{k=1}^{\infty}k^{-\alpha}(1-\varepsilon)^k=\text{Li}_{\alpha}(1-\epsilon)=$$ $$\qquad=\varepsilon^{\alpha} \left[\frac{\Gamma (1-\alpha)}{\varepsilon}+\tfrac{1}{2} (\alpha-1) \Gamma (1-\alpha)+\tfrac{1}{24} \left(3 \alpha^2-\alpha-2\right) \Gamma (1-\alpha) \varepsilon+O\left(\varepsilon^2\right)\right]$$ $$\qquad+\left[\zeta (\alpha)-\zeta (\alpha-1) \varepsilon+O\left(\varepsilon^2\right)\right].$$

Concerning the generalization, note that the finite sum $\sum_{k=1}^Na_k(1-\varepsilon)^k$ has for small $\varepsilon$ the expansion $b+c\varepsilon+O(\varepsilon^2)$, while $\sum_{k=N+1}^\infty k^{-\alpha}(1-\varepsilon)^k$ has the expansion $b'+\Gamma(1-\alpha)\varepsilon^{\alpha-1}+O(\varepsilon)$. So for $1<\alpha<2$ the large-$k$ behavior of $a_k$ dominates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.