Could anyone possibly give me any help with finding the minimum of this function? I believe the result to be $2\pi |n|$ from page 619 of this paper by W. G. C. Boyd.
\begin{equation} \frac{\frac{1}{2}(1+\zeta(m))\Gamma(m)}{(2\pi)^{m+1}|n|^m} \end{equation}
Thanks!
When $m$ is large, we have $$ a_m : = \frac{{\frac{1}{2}\left( {1 + \zeta \left( m \right)} \right)\Gamma \left( m \right)}}{{\left( {2\pi } \right)^{m + 1} \left| n \right|^m }} \sim \frac{{\Gamma \left( m \right)}}{{\left( {2\pi } \right)^{m + 1} \left| n \right|^m }} \sim \frac{{m^{m - 1/2} e^{ - m} \sqrt {2\pi } }}{{\left( {2\pi } \right)^{m + 1} \left| n \right|^m }} = \left( {\frac{m}{{2\pi e\left| n \right|}}} \right)^m \frac{1}{{\sqrt {2\pi m} }}. $$ Whence $$ \frac{a_m}{a_{m + 1}} \sim \frac{\left( \frac{m}{2\pi e\left| n \right|} \right)^m \frac{1}{{\sqrt {2\pi m} }}}{\left( \frac{{m + 1}}{{2\pi e\left| n \right|}} \right)^{m + 1} \frac{1}{\sqrt {2\pi \left( m + 1 \right)} }} \sim \frac{{\left( {\frac{m}{{2\pi e\left| n \right|}}} \right)^m }}{{\left( {\frac{{m + 1}}{{2\pi e\left| n \right|}}} \right)^{m + 1} }} = \frac{2\pi e\left| n \right|}{m + 1}\frac{1}{\left( 1 + \frac{1}{m} \right)^m } \sim \frac{2\pi e\left| n \right|}{m}\frac{1}{e} = \frac{2\pi \left| n \right|}{m}. $$ So for large $m$, your sequence attends its minimum at $m=2\pi \left| n \right|$.