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Question Number 116057 by Study last updated on 30/Sep/20

Σ_(n=2) ^∞ (3/(3n+1))=?

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{\mathrm{3}{n}+\mathrm{1}}=? \\ $$

Answered by mindispower last updated on 30/Sep/20

Σ_(n≥0) (((−1)^n )/((3n+1)))?

$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)}? \\ $$

Commented by mnjuly1970 last updated on 01/Oct/20

answer::= (π/(3(√3)))+((ln(2))/3) ✓

$${answer}::=\:\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 01/Oct/20

(1/2)Σ_(n=−∞) ^∞ (((−1)^n )/((3n+1))) ≠Σ_(n=0) ^∞  (((−1)^n )/(3n+1))  l.h.s=(1/2)[.....+(1/(14))−(1/(11)) +(1/8)−(1/5)   +(1/2) +(1/1)−(1/4)++(1/7)−(1/(10))+...

$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)}\:\neq\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}} \\ $$$${l}.{h}.{s}=\frac{\mathrm{1}}{\mathrm{2}}\left[.....+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{11}}\:+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{5}}\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}++\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{10}}+...\right. \\ $$$$ \\ $$$$\:\: \\ $$

Commented by mnjuly1970 last updated on 01/Oct/20

 because  Σ_(n=0) ^∞  (((−1)^n )/(3n+1)) =Σ_(n=0) ^∞ (−1)^n ∫_0 ^( 1) x^(3n) dx    =∫_0 ^( 1) (1/(1+x^3 )) dx=(π/(3(√3)))+((ln(2))/3)  ✓       .m.n.july.70

$$\:{because} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}} {dx}\:\: \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }\:{dx}=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark \\ $$$$\:\:\:\:\:.{m}.{n}.{july}.\mathrm{70} \\ $$$$\: \\ $$

Answered by mathmax by abdo last updated on 30/Sep/20

this serie is divergent due to (3/(3n+1))∼(1/n)

$$\mathrm{this}\:\mathrm{serie}\:\mathrm{is}\:\mathrm{divergent}\:\mathrm{due}\:\mathrm{to}\:\frac{\mathrm{3}}{\mathrm{3n}+\mathrm{1}}\sim\frac{\mathrm{1}}{\mathrm{n}} \\ $$

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