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As I was reading a book on the financial market micro-structure, I came across a simplification that I have not been able to prove.

The book states that $\sum_{\ell=1}^{Q}2G_0(\frac{1+\gamma}{\ell})\ell^\delta \approx 2G_0(1+\gamma)\frac{Q^\delta}{\delta}, \text{where } Q >> 1$ and $0<\delta<1$.

The first simplification is obvious: $\sum_{\ell=1}^{Q}2G_0(\frac{1+\gamma}{\ell})\ell^\delta = 2G_0(1+\gamma)\sum_{\ell=1}^{Q}\ell^{\delta-1}$.

I am not sure how $\sum_{\ell=1}^{Q}\ell^{\delta-1} \approx \frac{Q^\delta}{\delta}, \text{for where Q >> 1 and } 0<\delta<1$.

Any help will be extremely appreciated.

Arturo Magidin
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vpy
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1 Answers1

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$$\sum_{\ell=1}^{Q}\ell^{\delta-1}=H_Q^{(1-\delta )}$$ where appear the generalized harmonic numbers.

For large values of $Q$, we have $$H_Q^{(1-\delta )}=Q^{\delta -1} \left(\frac{Q}{\delta }+\frac{1}{2}+\frac{\delta -1}{12 Q}+O\left(\frac{1}{Q^3}\right)\right)+\zeta (1-\delta )$$

Claude Leibovici
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  • Thank you @Claude. Where did you get the formula for $H_Q^{1-\delta}$ for large values of Q from? – vpy May 06 '20 at 05:21
  • @vpy. Have a look at https://math.stackexchange.com/questions/1583452/assymptotics-of-the-generalized-harmonic-number-h-n-r-for-r-1 – Claude Leibovici May 06 '20 at 05:39