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Question Number 48283 by naka3546 last updated on 21/Nov/18

Commented by maxmathsup by imad last updated on 22/Nov/18

let S_n =Σ_(k=1) ^n (((−1)^(k+1) )/(2k−1)) changement of indice k−1=j give S_n =Σ_(j=0) ^(n−1) (((−1)^j )/(2j+1)) =Σ_(j=0) ^n (((−1)^j )/(2j+1)) −(((−1)^n )/(2n+1)) →(π/4) ⇒ −4S_n →−π⇒π−4S_(n/) →0 ⇒∃ a error a the Question..

letSn=∑k=1n(−1)k+12k−1changementofindicek−1=jgiveSn=∑j=0n−1(−1)j2j+1=∑j=0n(−1)j2j+1−(−1)n2n+1→π4⇒−4Sn→−π⇒π−4Sn→0⇒∃aerroratheQuestion..

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Nov/18

s=Σ_(k=1) ^n (((−1)^(k+1) )/(2k−1)) =(1/1)−(1/3)+(1/5)−...+(((−1)^(n+1) )/(2n−1)) if n→∞ s_∞ =(1/1)−(1/3)+(1/5)−...+(((−1)^(n+1) )/(2n−1))+...∞ gregory series... tan^(−1) x=x−(x^3 /3)+(x^5 /5)+... put x=1 (π/4)=1−(1/3)+(1/5)−(1/7)+... so answer is coming π−4×(π/4)=0

s=∑nk=1(−1)k+12k−1=11−13+15−...+(−1)n+12n−1ifn→∞s∞=11−13+15−...+(−1)n+12n−1+...∞gregoryseries...tan−1x=x−x33+x55+...putx=1π4=1−13+15−17+...soansweriscomingπ−4×π4=0

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