How do i show that $\prod_{k=m}^{n}(1+q/k) = \mathcal{O}(n^{q})$? I need to prove that
$\prod_{k=m}^{n}(1+q/k) = \mathcal{O}(n^{q})$, as functions of $n$, with $q < 0$ and $0 <m \leq n$.
Anyone has an idea how to do this?
I thought about showing that $\sum_{k=m}^{n}\log(1-k/q)/(q\log(n))$ converges for increasing $n$ but i got stuck there.
 A: Your idea is good. You can use the fact that $\log(1+q/k)$ is increasingly well approximated by $q/k$ as $k \to \infty$.
We have the following equalities.
\begin{align}
\lim_{n \to \infty} \frac{\sum_{k=m}^{n} \log(1+q/k)}{q \cdot \log(n)} \approx& \lim_{n \to \infty} \frac{\sum_{k=m}^n \frac{q}{k}}{q \log (n)} \\
=& \lim_{n \to \infty} \frac{\sum_{k=m}^n \frac{1}{k}}{\log (n)}
\end{align}
And this converges since the difference between the harmonic series and the logarithm approaches the Euler-Masceroni constant.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\bbox[5px,#ffd]{\prod_{k\ =\ m}^{n}
\pars{1 + {q \over k}} = \mathcal{O}\pars{n^{q}}}:\
{\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\prod_{k\ =\ m}^{n}
\pars{1 + {q \over k}}} =
\prod_{k\ =\ m}^{n}{k + q \over k} =
{\pars{m + q}^{\overline{n - m + 1}} \over
m^{\overline{n - m + 1}}}
\\[5mm] = &\
{\Gamma\pars{q + n + 1}/\Gamma\pars{m + q} \over
\Gamma\pars{n + 1}/\Gamma\pars{m}} =
{\Gamma\pars{m} \over \Gamma\pars{m + q}}{\pars{n + q}! \over n!}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
{\Gamma\pars{m} \over \Gamma\pars{m + q}}
{\root{2\pi}\pars{n + q}^{n + q + 1/2}\,\,\,
\expo{-n - q} \over \root{2\pi}n^{n + 1/2}\,\,\,
\expo{-n}}
\\[5mm] = &\
{\Gamma\pars{m} \over \Gamma\pars{m + q}}
{n^{n + q + 1/2}\,\,\,\pars{1 + q/n}^{n + q + 1/2}
 \over n^{n + 1/2}}\,\,\expo{-q}
\\[5mm] = &\
{\Gamma\pars{m} \over \Gamma\pars{m + q}}\,n^{q}\,\
\underbrace{\bracks{\pars{1 + {q \over n}}^{n + q + 1/2}\,\,\,
\expo{-q}}}_{\ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\quad 1}}
\end{align}

\begin{align}
&\bbx{\bbox[5px,#ffd]{\prod_{k\ =\ m}^{n}
\pars{1 + {q \over k}}} \sim
{\Gamma\pars{m} \over \Gamma\pars{m + q}}\,
\color{red}{n^{q}}\quad\mrm{as}\quad n \to \infty}
\\ &
\end{align}
