Menu Close

calculate-n-1-1-n-n-2n-1-2-

Tinku Tara 0 Comments
Pin




Question Number 68593 by Abdo msup. last updated on 14/Sep/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)^2 ))
$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by Abdo msup. last updated on 06/Oct/19
let S=Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)^2 )) first let decompose F(x)=(1/(x(2x+1)^2 )) ⇒F(x)=(a/x) +(b/(2x+1)) +(c/((2x+1)^2 )) a=lim_(x→0) xF(x)=1 c=lim_(x→−(1/2)) (2x+1)^2 F−x)=−2 ⇒ F(x)=(1/x) +(b/(2x+1))−(2/((2x+1)^2 )) lim_(x→+∞) xF(x)=0 =1+(b/2) ⇒b+2=0 ⇒b=−2 ⇒ F(x)=(1/x) −(2/(2x+1)) −(2/((2x+1)^2 )) ⇒ S=Σ_(n=1) ^∞ (((−1)^n )/n) −2Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) −2Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) we have Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/((2n+1))) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) −1=(π/4)−1 rest to ca<culate α_0 =Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) ⇒ S=−ln(2)−(π/2) +2 −2α_0
$${let}\:{S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{first}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{{c}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)=\mathrm{1} \\ $$$$\left.{c}={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} {F}−{x}\right)=−\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{\mathrm{2}{x}+\mathrm{1}}−\frac{\mathrm{2}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\mathrm{0}\:=\mathrm{1}+\frac{{b}}{\mathrm{2}}\:\Rightarrow{b}+\mathrm{2}=\mathrm{0}\:\Rightarrow{b}=−\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\:−\frac{\mathrm{2}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}}\:−\mathrm{1}=\frac{\pi}{\mathrm{4}}−\mathrm{1} \\ $$$${rest}\:{to}\:{ca}<{culate}\:\alpha_{\mathrm{0}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}=−{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}\:−\mathrm{2}\alpha_{\mathrm{0}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *