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Question Number 156776 by mnjuly1970 last updated on 15/Oct/21
Σ_(n=1) ^∞ (n ln(((2n+1)/(2n−1))) −1)=?
$$ \\ $$$$\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}\:{ln}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right)\:−\mathrm{1}\right)=? \\ $$$$ \\ $$
Commented by CAIMAN last updated on 15/Oct/21
1/2-1/2 ln2
Answered by qaz last updated on 15/Oct/21
Σ_(n=1) ^∞ (nln((2n+1)/(2n−1))−1) =Σ_(n=1) ^∞ (nln(2n+1)−nln(2n−1)−1) =Σ_(n=1) ^∞ ∫_0 ^1 ((n/(x+2n))−(n/(x−2n))−1)dx =∫_0 ^1 Σ_(n=1) ^∞ ((n/(x+2n))−(n/(x−2n))−1)dx =(1/4)∫_0 ^1 x^2 Σ_(n=1) ^∞ (1/(n^2 −((x/2))^2 ))dx =(1/4)∫_0 ^1 x^2 ((1−π((x/2))cot ((πx)/2))/(2((x/2))^2 ))dx =(1/4)∫_0 ^1 (2−πxcot ((πx)/2))dx =(1/4)(2−2ln2) =(1/2)(1−ln2)
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{nln}\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{2n}−\mathrm{1}}−\mathrm{1}\right) \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{nln}\left(\mathrm{2n}+\mathrm{1}\right)−\mathrm{nln}\left(\mathrm{2n}−\mathrm{1}\right)−\mathrm{1}\right) \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{n}}{\mathrm{x}+\mathrm{2n}}−\frac{\mathrm{n}}{\mathrm{x}−\mathrm{2n}}−\mathrm{1}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{n}}{\mathrm{x}+\mathrm{2n}}−\frac{\mathrm{n}}{\mathrm{x}−\mathrm{2n}}−\mathrm{1}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}} \frac{\mathrm{1}−\pi\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cot}\:\frac{\pi\mathrm{x}}{\mathrm{2}}}{\mathrm{2}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}−\pi\mathrm{xcot}\:\frac{\pi\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}−\mathrm{2ln2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{ln2}\right) \\ $$
Commented by mnjuly1970 last updated on 15/Oct/21
bravo sir qaz
$${bravo}\:{sir}\:{qaz} \\ $$

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