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Question Number 43259 by maxmathsup by imad last updated on 08/Sep/18
calculate Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1))
$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 09/Sep/18
let S = Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1)) ⇒S=(1/2) Σ_(n=2) ^∞ (−1)^n {(1/(n−1)) −(1/(n+1))} ⇒ 2S = Σ_(n=2) ^∞ (((−1)^n )/(n−1)) −Σ_(n=2) ^∞ (((−1)^n )/(n+1)) but Σ_(n=2) ^∞ (((−1)^n )/(n−1)) = Σ_(n=1) ^∞ (((−1)^(n+1) )/n) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(2) (Σ_(n=1) ^∞ (x^n /n) = −ln(1−x) with −1<x<1) also Σ_(n=2) ^∞ (((−1)^n )/(n+1)) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −{1 −(1/2)} =ln(2) −(1/2) ⇒ 2S = ln(2) −ln(2)+(1/2) ⇒ S =(1/4) .
$${let}\:{S}\:=\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{S}=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left\{\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\} \\ $$$$\:\Rightarrow\:\mathrm{2}{S}\:=\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:−\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:\:{but}\:\: \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:={ln}\left(\mathrm{2}\right) \\ $$$$\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}}\:=\:−{ln}\left(\mathrm{1}−{x}\right)\:\:{with}\:−\mathrm{1}<{x}<\mathrm{1}\right)\:\:{also} \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} }{\boldsymbol{{n}}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:−\left\{\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$={ln}\left(\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow\:\mathrm{2}{S}\:=\:{ln}\left(\mathrm{2}\right)\:−{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow\:{S}\:=\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$

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