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Question Number 69212 by aliesam last updated on 21/Sep/19

$$uselimitcomparisontesttodetermine$$$$theseriesconvergeordiverge$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}ln(1+\frac{1}{k^2})$$

Answered by mind is power last updated on 21/Sep/19

$$\mathrm{\forall }x\in {IR}_{+}\ln (1+x)⩽x$$$$\Rightarrow \mathrm{\forall }k\in {\mathbb{N}}^{\ast }0⩽\ln (1+\frac{1}{k^2})⩽\frac{1}{k^2}$$$$\Rightarrow \underset{k=1}{\sum ^{+\mathrm{\infty }}}\ln (1+\frac{1}{k^2})⩽\sum _{k=1}^{+\mathrm{\infty }}\frac{1}{k^2}=\frac{{\pi }^2}{6}$$

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