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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

$${prove}\:{that}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\:{ln}\mathrm{2} \\ $$

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))

$${f}\left({x}\right)\:=\:\mathrm{ln}\left({x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:{f}'\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:{f}''\left({x}\right)\:=\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{f}'''\left({x}\right)\:=\:\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:{f}''''\left({x}\right)\:=\:−\frac{\mathrm{6}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\vdots \\ $$$$\:\:\:\:\:\:{f}^{\left({k}\right)} \left({x}\right)\:=\:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \frac{\left({k}−\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)^{{k}} } \\ $$$$ \\ $$$${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{\left({k}\right)} \left({a}\right)}{{k}!}\left({x}−{a}\right)^{{k}} \\ $$$${a}\:=\:\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({a}\right)\:=\:\mathrm{0}}\\{{f}^{\left({k}\right)} \left({a}\right)\:=\:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}\end{cases}\:\:\:\therefore\:\:{f}\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{{k}!}{x}^{{k}} \:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {x}^{{k}} }{{k}} \\ $$$${k}\:=\:{n}+\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\mathrm{ln}\left(\mathrm{2}\right)\:=\:{f}\left(\mathrm{1}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$$$\:\:\: \\ $$

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))))

$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx} \\ $$$$\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{x}\right)^{{n}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{n}} {dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{dx}=\left[\mathrm{ln}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\Rightarrow\begin{array}{|c|}{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}=\mathrm{ln}\left(\mathrm{2}\right)}\\\hline\end{array} \\ $$

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