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Question Number 162424 by mnjuly1970 last updated on 29/Dec/21

calculate Ω = Σ_(n=1) ^∞ (( (−1)^( n) n)/(3^( n) (2n −1 ))) =? − Inspired from Sir Ghaderi′s post−

calculateΩ=∑∞n=1(−1)nn3n(2n−1)=?−InspiredfromSirGhaderi′spost−

Answered by Ar Brandon last updated on 29/Dec/21

Ω=Σ_(n=1) ^∞ (((−1)^n n)/(3^n (2n−1)))=(1/2)Σ_(n=1) ^∞ (−(1/3))^n +(1/2)Σ_(n=1) ^∞ (−(1/3))^n (1/(2n−1)) =−(1/8)+(1/2)Σ_(n=1) ^∞ (−(1/3))^n ∫_0 ^1 x^(2n−2) dx=−(1/8)+(1/2)∫_0 ^1 (1/x^2 )Σ_(n=1) ^∞ (−(x^2 /3))^n dx =−(1/8)−(1/2)∫_0 ^1 (1/x^2 )∙((x^2 /3)/(1+(x^2 /3)))dx=−(1/8)−(1/2)∫_0 ^1 (1/(x^2 +3))dx =−(1/8)−[(1/(2(√3)))arctan((x/( (√3))))]_0 ^1 =−(1/8)−(π/(12(√3)))

Ω=∑∞n=1(−1)nn3n(2n−1)=12∑∞n=1(−13)n+12∑∞n=1(−13)n12n−1=−18+12∑∞n=1(−13)n∫01x2n−2dx=−18+12∫011x2∑∞n=1(−x23)ndx=−18−12∫011x2⋅x231+x23dx=−18−12∫011x2+3dx=−18−[123arctan(x3)]01=−18−π123

Commented by mnjuly1970 last updated on 29/Dec/21

thank you so much sir brandon

thankyousomuchsirbrandon

Commented by Ar Brandon last updated on 29/Dec/21

My pleasure, Sir !

Mypleasure,Sir!

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