Asymptotic of a series with power-law decaying summands Maybe someone has an idea for the following: Let us fix $i\in\mathbb N$ and $\alpha\in \left(0,\frac12\right)$.
My question is, how does
$$\sum_{m=1}^n m^{\alpha} \left( m+i \right)^{\alpha -1}$$
behave for $n\rightarrow \infty$?
My guess would be that there exists a constant $C_{i,\alpha}$ depending only on $i$ and $\alpha$ such that
$$ \sum_{m=1}^n m^{\alpha} \left( m+i \right)^{\alpha -1} \sim C_{i, \alpha} n^{2\alpha}.$$
Does anyone see some straight way to proof that?
And does anyone see what $C_{i, \alpha}$ might be? It must at least contain something like $\frac{1}{2\alpha}$, I believe?
All the best and thank you all!
 A: Fix $i\in \mathbb{N}$. Then
\begin{align*}
& \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sum\limits_{m = 1}^{n + 1} {m^\alpha  (m + i)^{\alpha  - 1} }  - \sum\limits_{m = 1}^n {m^\alpha  (m + i)^{\alpha  - 1} } }}{{(n + 1)^{2\alpha }  - n^{2\alpha } }} \\ & = \mathop {\lim }\limits_{n \to  + \infty } \frac{{(n + 1)^\alpha  (n + 1 + i)^{\alpha  - 1} }}{{(n + 1)^{2\alpha }  - n^{2\alpha } }}
\\ &  = \mathop {\lim }\limits_{n \to  + \infty } \frac{{(n + 1)^\alpha  (n + 1 + i)^{\alpha  - 1} }}{{((n + 1)^\alpha   + n^\alpha  )((n + 1)^\alpha   - n^\alpha  )}} \\ &= \mathop {\lim }\limits_{n \to  + \infty } \frac{{(n + 1)^\alpha  (n + 1 + i)^{\alpha  - 1} }}{{((n + 1)^\alpha   + n^\alpha  )(n^\alpha   + \alpha n^{\alpha  - 1}  + \mathcal{O}\!\left( {n^{\alpha  - 2} } \right) - n^\alpha  )}} = \frac{1}{{2\alpha }}.
\end{align*}
Thus, by the Stolz–Cesàro theorem,
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{{\sum\limits_{m = 1}^n {m^\alpha  (m + i)^{\alpha  - 1} } }}{{n^{2\alpha } }} = \frac{1}{{2\alpha }}.
$$
