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Question Number 33304 by abdo imad last updated on 14/Apr/18
$${find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{ln}\left(\:\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$
Commented by abdo imad last updated on 19/Apr/18
$${let}\:{put}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{{k}}\right){we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=_{{n}\rightarrow+\infty} {lim}\:{S}_{{n}} \\ $$$${S}_{{n}} ={ln}\left(\prod_{{k}=\mathrm{1}} ^{{n}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)\right)\:{but} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)=\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}+\mathrm{1}}{{k}}\:=\frac{\mathrm{2}}{\mathrm{1}}.\:\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{3}}\:....\frac{{n}}{{n}−\mathrm{1}}\:\frac{{n}+\mathrm{1}}{{n}}\:={n}+\mathrm{1} \\ $$$$\Rightarrow\:{S}_{{n}} ={ln}\left({n}+\mathrm{1}\right)\Rightarrow\:{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =+\infty\:. \\ $$