Question Number 213648 by issac last updated on 12/Nov/24
$${a}_{{m}} =\underset{{h}={m}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\underset{{h}=\mathrm{1}} {\overset{{m}} {\sum}}\:\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{2}} \\ $$$${a}_{{m}} \approx\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right) \\ $$$$\boldsymbol{\psi}^{\left({n}\right)} \left({z}\right)=\frac{\mathrm{d}^{{n}+\mathrm{1}} \:\:}{\mathrm{d}{z}^{{n}+\mathrm{1}} }\mathrm{ln}\left(\boldsymbol{\Gamma}\left({z}\right)\right)\:,\:\mathbb{C}\backslash\mathbb{Z}_{\leq\mathrm{0}} \\ $$$$\mathrm{aka}\:\:\mathrm{polygamma}\:\mathrm{function} \\ $$$$\mathrm{1}.\:\:\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:{m}\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)=\mathrm{1}. \\ $$$$\mathrm{and}… \\ $$$$\mathrm{2}.\:\:\:\underset{{m}\rightarrow\infty} {\mathrm{lim}}{m}^{\mathrm{2}} \boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)=\infty\:\mathrm{div} \\ $$$${m}^{\mathrm{2}} \boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)\approx{m}−\boldsymbol{\varepsilon} \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:{m}^{\mathrm{2}} \boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)\approx\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:\left({m}−\boldsymbol{\varepsilon}\right)=\infty \\ $$