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Question Number 41407 by maxmathsup by imad last updated on 06/Aug/18
calculate Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)(n+2)))
calculate∑n=1∞(−1)nn(n+1)(n+2)
Commented by maxmathsup by imad last updated on 08/Aug/18
let f(x)= Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)(n+2))) x^(n+2) with ∣x∣<1 we have f^′ (x)=Σ_(n=1) ^∞ (((−1)^n )/(n(n+1))) x^(n+1) =Σ_(n=1) ^∞ (−1)^n {(1/n) −(1/(n+1))}x^(n+1) =Σ_(n=1) ^∞ (((−1)^n )/n) x^(n+1) −Σ_(n=1) ^∞ (((−1)^n )/(n+1)) x^(n+1) but Σ_(n=1) ^∞ (((−1)^n )/n) x^(n+1) =−x Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n =−xln(1+x) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) x^(n+1) = Σ_(n=2) ^∞ (((−1)^(n−1) )/n) x^n =ln(1+x)−1 ⇒ f^′ (x)=−xln(1+x)−ln(1+x) +1 ⇒f(x)= ∫_0 ^x {−(t+1)ln(1+t) +1}dt +c c=f(o)=0 ⇒ f(x)=x− ∫_0 ^x (t+1)ln(t+1)dt but ∫_0 ^x (t+1)ln(t+1)dt[=_(t+1 =u) ∫_1 ^(x+1) u ln(u)du =[(u^2 /2)ln(u)]_1 ^(x+1) −∫_1 ^(x+1) (u/2) du = (((x+1)^2 )/2)ln(x+1) −(1/4)[u^2 ]_1 ^(x+1) =(((x+1)^2 )/2)ln(x+1)−(1/4)(x+1)^2 +(1/4) ⇒ f(x)=x −(((x+1)^2 )/2)ln(x+1) +(((x+1)^2 )/4) −(1/4) S=f(1) =1−2ln(2) +1−(1/4) =−2ln(2) +2−(1/4) S =−2ln(2) +(7/4)
letf(x)=∑n=1∞(−1)nn(n+1)(n+2)xn+2with∣x∣<1wehavef′(x)=∑n=1∞(−1)nn(n+1)xn+1=∑n=1∞(−1)n{1n−1n+1}xn+1=∑n=1∞(−1)nnxn+1−∑n=1∞(−1)nn+1xn+1but∑n=1∞(−1)nnxn+1=−x∑n=1∞(−1)n−1nxn=−xln(1+x)∑n=1∞(−1)nn+1xn+1=∑n=2∞(−1)n−1nxn=ln(1+x)−1⇒f′(x)=−xln(1+x)−ln(1+x)+1⇒f(x)=∫0x{−(t+1)ln(1+t)+1}dt+cc=f(o)=0⇒f(x)=x−∫0x(t+1)ln(t+1)dtbut∫0x(t+1)ln(t+1)dt[=t+1=u∫1x+1uln(u)du=[u22ln(u)]1x+1−∫1x+1u2du=(x+1)22ln(x+1)−14[u2]1x+1=(x+1)22ln(x+1)−14(x+1)2+14⇒f(x)=x−(x+1)22ln(x+1)+(x+1)24−14S=f(1)=1−2ln(2)+1−14=−2ln(2)+2−14S=−2ln(2)+74

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