$$\sum\limits_{k=1}^n \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right),$$ where $0<\alpha<2$ and $\alpha\neq1$. Can someone help me in finding the asymptotic behavior of finite sum?
I already have the asymptotic behavior of this finite sum for $\alpha=1$.
I do not know how we could make it for any value of $\alpha$.
However, we can notice that, if $$S_n^{(\alpha)}=\sum\limits_{k=1}^n \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right)$$ then $2^\alpha S_n^{(\alpha)}$ can write $$ -\zeta \left(\alpha ,1+2^{\frac{1}{\alpha }-1}\right)+\zeta \left(\alpha ,\frac{1}{2} \left(1+2^{\frac{1}{\alpha }}\right)\right)+\zeta \left(\alpha ,n+2^{\frac{1}{\alpha }-1}+1\right)-\zeta \left(\alpha ,\frac{1}{2} \left(1+2^{\frac{1}{\alpha }}\right)+n\right)$$ where appears the Hurwitz zeta function.
For $\alpha=2$, this would give $$S_n^{(2)}=\frac{1}{4} \left(\zeta \left(2,\frac{1}{2} \left(1+\sqrt{2}\right)\right)-\zeta \left(2,1+\frac{1}{\sqrt{2}}\right)\right)-\frac{1}{8 n^2}+\frac{\frac{1}{16}+\frac{1}{4 \sqrt{2}}}{n^3}+O\left(\frac{1}{n^4}\right)$$ the constant coefficient being $\approx 0.116998$.
As shown in the table below, this seems to be quite good $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 5 & 0.1134940446 & 0.1139123171 \\ 6 & 0.1144254293 & 0.1146336438 \\ 7 & 0.1150296641 & 0.1151446828 \\ 8 & 0.1154436961 & 0.1155123159 \\ 9 & 0.1157396829 & 0.1157831196 \\ 10 & 0.1159585608 & 0.1159873803 \end{array} \right)$$
It is interesting to notice that this is quite workable for integer values of $\alpha$ leading to something like $$S_n^{(\alpha)}=C_\alpha-\frac 1 {2^{\alpha+1} n^\alpha}+O\left(\frac{1}{n^{\alpha+1}}\right)$$
May be, this could extend to non integer values of $\alpha$.