Question Number 69212 by aliesam last updated on 21/Sep/19
$${use}\:{limit}\:{comparison}\:{test}\:{to}\:{determine} \\ $$$${the}\:{series}\:{converge}\:{or}\:{diverge} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$
Answered by mind is power last updated on 21/Sep/19
$$\forall{x}\in{IR}_{+} \:\:\:\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\leqslant{x} \\ $$$$\Rightarrow\:\forall{k}\in\mathbb{N}^{\ast} \:\:\mathrm{0}\leqslant\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\leqslant\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\leqslant\sum_{{k}=\mathrm{1}} ^{+\infty} \frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$