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find-n-1-ln-1-1-n-




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

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