Question Number 145270 by mnjuly1970 last updated on 03/Jul/21
$$ \\ $$$$\:\:\:\:\:\:\:#\:\mathrm{Calculus}\:\left(\:\mathrm{I}\:\right)\:# \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{Arccot}\left(\mathrm{3}\:+\frac{{n}\:\left(\:{n}\:+\:\mathrm{1}\right)}{\mathrm{3}}\:\right)=\:? \\ $$$$\:\:\:\:\:\:….. \\ $$
Answered by Olaf_Thorendsen last updated on 03/Jul/21
$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{Arccot}\left(\mathrm{3}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{3}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{Arctan}\left(\frac{\mathrm{1}}{\mathrm{3}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{3}}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{Arctan}\left(\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{9}}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{Arctan}\left(\frac{\frac{{n}+\mathrm{1}}{\mathrm{3}}−\frac{{n}}{\mathrm{3}}}{\mathrm{1}+\frac{{n}}{\mathrm{3}}×\frac{\left({n}+\mathrm{1}\right)}{\mathrm{3}}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{Arctan}\left(\frac{{n}+\mathrm{1}}{\mathrm{3}}\right)−\mathrm{Arctan}\left(\frac{{n}}{\mathrm{3}}\right)\right) \\ $$$$\mathrm{S}\:=\:\frac{\pi}{\mathrm{2}}−\mathrm{Arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\mathrm{S}\:=\:\frac{\pi}{\mathrm{2}}−\left(\frac{\pi}{\mathrm{2}}−\mathrm{Arctan}\left(\mathrm{3}\right)\right) \\ $$$$\mathrm{S}\:=\:\mathrm{Arctan}\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 03/Jul/21
$$\:\:\:\:{very}\:{nice}\:{mr}\:{olaf}… \\ $$