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Question Number 90562 by Tony Lin last updated on 24/Apr/20

prove that Σ_(n=1) ^∞ (1/(2n(2n−1)))=ln2

$${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}={ln}\mathrm{2} \\ $$

Commented by mathmax by abdo last updated on 24/Apr/20

let S_n =Σ_(k=1) ^n (1/(2k(2k−1))) ⇒ S_n =Σ_(k=1) ^n ((1/(2k−1))−(1/(2k))) =Σ_(k=1) ^n (1/(2k−1))−(1/2)Σ_(k=1) ^n (1/k) we have Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/(2k−1)) =1+(1/3)+(1/5)+...+(1/(2n−1)) =1+(1/2)+(1/3)+...+(1/(2n−1))+(1/(2n)) −(1/2)−(1/4)−....−(1/(2n)) =H_(2n) −(1/2)H_n ⇒ S_n =H_(2n) −(1/2)H_n −(1/2)H_n =H_(2n) −H_n =ln(2n)+γ +o((1/(2n))) −ln(n)−γ−o((1/n)) =ln(2)+o((1/(2n)))−o((1/n))→ln(2) (n→+∞) ⇒ Σ_(n=1) ^∞ (1/(2n(2n−1))) =lim_(n→+∞) S_n =ln(2)

$${let}\:{S}_{{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}\left(\mathrm{2}{k}−\mathrm{1}\right)}\:\Rightarrow\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−....−\frac{\mathrm{1}}{\mathrm{2}{n}}\:={H}_{\mathrm{2}{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} \:\Rightarrow \\ $$$${S}_{{n}} ={H}_{\mathrm{2}{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} ={H}_{\mathrm{2}{n}} −{H}_{{n}} ={ln}\left(\mathrm{2}{n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$−{ln}\left({n}\right)−\gamma−{o}\left(\frac{\mathrm{1}}{{n}}\right)\:={ln}\left(\mathrm{2}\right)+{o}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)−{o}\left(\frac{\mathrm{1}}{{n}}\right)\rightarrow{ln}\left(\mathrm{2}\right)\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} ={ln}\left(\mathrm{2}\right) \\ $$

Commented by Tony Lin last updated on 25/Apr/20

thanks sir

$${thanks}\:{sir} \\ $$

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