Question Number 150654 by mnjuly1970 last updated on 14/Aug/21
Answered by Ar Brandon last updated on 14/Aug/21
![S=Σ_(n=1) ^∞ (((−1)^n )/((3n+1)(3n+2)))=Σ_(n=1) ^∞ (−1)^n ((1/(3n+1))−(1/(3n+2))) =Σ_(n=1) ^∞ (−1)^n ∫_0 ^1 (x^(3n) −x^(3n+1) )dx=∫_0 ^1 Σ_(n≥1) ((−x)^(3n) +(−x)^(3n+1) )dx =∫_0 ^1 (((−x^3 )/(1+x^3 ))+(x^4 /(1+x^3 )))dx=∫_0 ^1 ((x^3 (x−1))/(x^3 +1))dx=∫_0 ^1 (x−1−((x−1)/(x^3 +1)))dx =∫_0 ^1 (x−1−((x−1)/((x+1)(x^2 −x+1))))dx=∫_0 ^1 (x−1+(2/(3(x+1)))−((2x−1)/(3(x^2 −x+1))))dx =[(x^2 /2)−x+(2/3)ln(x+1)−(1/3)ln(x^2 −x+1)]_0 ^1 =−(1/2)+(2/3)ln(2)](Q150669.png)
$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}\right) \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}{n}} −{x}^{\mathrm{3}{n}+\mathrm{1}} \right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{1}} {\sum}\left(\left(−{x}\right)^{\mathrm{3}{n}} +\left(−{x}\right)^{\mathrm{3}{n}+\mathrm{1}} \right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{−{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{3}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{3}} }\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)}{{x}^{\mathrm{3}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−\mathrm{1}−\frac{{x}−\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−\mathrm{1}−\frac{{x}−\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right){dx} \\ $$$$\:\:\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$
Commented by Tawa11 last updated on 14/Aug/21
$$\mathrm{Great} \\ $$
Commented by mnjuly1970 last updated on 14/Aug/21
$$\:{grateful}..\:{very}\:{nice}... \\ $$