Question Number 119192 by help last updated on 22/Oct/20
Answered by Olaf last updated on 22/Oct/20
$$\forall{k}\in\mathbb{N}^{\ast} ,\:{k}^{\mathrm{2}} +\mathrm{1}\:>\:{k}^{\mathrm{2}} \\ $$$$\forall{k}\in\mathbb{N}^{\ast} ,\:\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \:>\:{k}^{\mathrm{6}} \\ $$$$\forall{k}\in\mathbb{N}^{\ast} ,\:\frac{\mathrm{1}}{\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:<\:\frac{\mathrm{1}}{{k}^{\mathrm{6}} } \\ $$$$\forall{k}\in\mathbb{N}^{\ast} ,\:\frac{{k}}{\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:<\:\frac{\mathrm{1}}{{k}^{\mathrm{5}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}}{\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:<\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{5}} } \\ $$$$\mathrm{And}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{5}} }\:\mathrm{converges}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{5}} }\:=\:\zeta\left(\mathrm{5}\right)\:\approx\:\mathrm{1},\mathrm{037} \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}}{\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{converges}. \\ $$$$ \\ $$