Author: Tinku Tara

Question Number 203774 by Calculusboy last updated on 27/Jan/24 $$\boldsymbol{{determine}}\:\boldsymbol{{whether}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{convergent}}\:\boldsymbol{{or}}\:\boldsymbol{{divergent}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{{n}}}{\:\sqrt{\mathrm{4}\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}} \\ $$ Answered by witcher3 last updated on…

Question Number 203771 by Calculusboy last updated on 27/Jan/24 Answered by DwaipayanShikari last updated on 27/Jan/24 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\begin{pmatrix}{{n}+\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!}{\left({n}+\mathrm{3}\right)!\mathrm{3}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}\underset{{n}=\mathrm{0}}…