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Question Number 161623 by amin96 last updated on 20/Dec/21
Σ_(n=1) ^∞ (((−1)^(n+1) )/(n(n+2)))=?
∑∞n=1(−1)n+1n(n+2)=?
Answered by TheSupreme last updated on 20/Dec/21
(A/n)+(B/(n+2))=(((−1)^(n+1) )/(n(n+2))) An+2A+Bn=(−1)^(n+1) A=−B=(((−1)^(n+1) )/2) Σ_(n=1) ^∞ (((−1)^(n+1) )/2)(1/n)−(((−1)^(n+1) )/2)(1/(n+2)) s_n =(1/2)−(1/3)+(((−1)^n )/(n+2)) lim s_n =(1/2)−(1/3)=(1/6)
An+Bn+2=(−1)n+1n(n+2)An+2A+Bn=(−1)n+1A=−B=(−1)n+12∑∞n=1(−1)n+121n−(−1)n+121n+2sn=12−13+(−1)nn+2limsn=12−13=16
Commented by mr W last updated on 21/Dec/21
how did you come from An+2A+Bn=(−1)^(n+1) to A=−B=(((−1)^(n+1) )/2) ?
howdidyoucomefromAn+2A+Bn=(−1)n+1toA=−B=(−1)n+12?
Answered by mathmax by abdo last updated on 20/Dec/21
S_n =Σ_(k=1) ^n (((−1)^(k+1) )/(k(k+2))) ⇒S_n =(1/2)Σ_(k=1) ^n ((1/k)−(1/(k+2)))(−1)^(k+1) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k) +(1/2)Σ_(k=1) ^(n ) (((−1)^k )/(k+2)) (→k+2=p) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k)+(1/2)Σ_(p=3) ^(n+2) (((−1)^p )/p) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k)+(1/2){Σ_(p=1) ^n (((−1)^p )/p)−(−1+(1/2))+(((−1)^n )/(n+2))} =(1/2){(1/2)+(((−1)^n )/(n+2))} ⇒lim_(n→+∞) S_n =(1/4)
Sn=∑k=1n(−1)k+1k(k+2)⇒Sn=12∑k=1n(1k−1k+2)(−1)k+1=−12∑k=1n(−1)kk+12∑k=1n(−1)kk+2(→k+2=p)=−12∑k=1n(−1)kk+12∑p=3n+2(−1)pp=−12∑k=1n(−1)kk+12{∑p=1n(−1)pp−(−1+12)+(−1)nn+2}=12{12+(−1)nn+2}⇒limn→+∞Sn=14

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