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Question Number 161755 by mkam last updated on 22/Dec/21

prove that Σ_(n=0) ^∞ (((−1)^n )/(n+1)) = ln2

provethat∑∞n=0(−1)nn+1=ln2

Answered by FelipeLz last updated on 22/Dec/21

f(x) = ln(x+1) f′(x) = (1/(x+1)) f′′(x) = −(1/((x+1)^2 )) f′′′(x) = (2/((x+1)^3 )) f′′′′(x) = −(6/((x+1)^4 )) ⋮ f^((k)) (x) = (−1)^(k−1) (((k−1)!)/((x+1)^k )) f(x) = Σ_(k=0) ^∞ ((f^((k)) (a))/(k!))(x−a)^k a = 0 ⇒ { ((f(a) = 0)),((f^((k)) (a) = (−1)^(k−1) (k−1)!)) :} ∴ f(x) = Σ_(k=1) ^∞ (((−1)^(k−1) (k−1)!)/(k!))x^k = Σ_(k=1) ^∞ (((−1)^(k−1) x^k )/k) k = n+1 ⇒ f(x) = Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) ln(2) = f(1) = Σ_(n=0) ^∞ (((−1)^n (1)^(n+1) )/(n+1)) = Σ_(n=0) ^∞ (((−1)^n )/(n+1))

f(x)=ln(x+1)f′(x)=1x+1f″(x)=−1(x+1)2f‴(x)=2(x+1)3f⁗(x)=−6(x+1)4⋮f(k)(x)=(−1)k−1(k−1)!(x+1)kf(x)=∑∞k=0f(k)(a)k!(x−a)ka=0⇒{f(a)=0f(k)(a)=(−1)k−1(k−1)!∴f(x)=∑∞k=1(−1)k−1(k−1)!k!xk=∑∞k=1(−1)k−1xkkk=n+1⇒f(x)=∑∞n=0(−1)nxn+1n+1ln(2)=f(1)=∑∞n=0(−1)n(1)n+1n+1=∑∞n=0(−1)nn+1

Answered by Ar Brandon last updated on 22/Dec/21

S=Σ_(n=0) ^∞ (((−1)^n )/(n+1))=Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n dx =Σ_(n=0) ^∞ ∫_0 ^1 (−x)^n dx=∫_0 ^1 Σ_(n=0) ^∞ (−x)^n dx =∫_0 ^1 (1/(1+x))dx=[ln(1+x)]_0 ^1 =ln(2)−ln(1) ⇒ determinant (((Σ_(n=0) ^∞ (((−1)^n )/(n+1))=ln(2))))

S=∑∞n=0(−1)nn+1=∑∞n=0(−1)n∫01xndx=∑∞n=0∫01(−x)ndx=∫01∑∞n=0(−x)ndx=∫0111+xdx=[ln(1+x)]01=ln(2)−ln(1)⇒∑∞n=0(−1)nn+1=ln(2)

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