n-0-1-n-2n-1-1-pi-4-2-pi-2-12-3-pi-8-4-pi-2-6-

Question Number 53620 by kaivan.ahmadi last updated on 23/Jan/19

Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) =? 1. (π/4) 2. (π^2 /(12)) 3. (π/8) 4. (π^2 /6)

$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}\:=? \\ $$$$ \\ $$$$\mathrm{1}.\:\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:\:\:\mathrm{2}.\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\:\:\:\:\mathrm{3}.\:\frac{\pi}{\mathrm{8}}\:\:\:\:\:\mathrm{4}.\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Commented by turbo msup by abdo last updated on 24/Jan/19

1)(π/4) if you want method see the platform.

$$\left.\mathrm{1}\right)\frac{\pi}{\mathrm{4}}\:\:{if}\:{you}\:{want}\:{method}\:{see}\:{the} \\ $$$${platform}. \\ $$

Commented by kaivan.ahmadi last updated on 24/Jan/19

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by turbo msup by abdo last updated on 24/Jan/19

$${you}\:{are}\:{welcome} \\ $$

Commented by maxmathsup by imad last updated on 26/Jan/19

let S(x) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(2n+1) with ∣x∣<1 due to uniform convergence we have (d/dx)S(x) =Σ_(n=0) ^∞ (−1)^n x^(2n) =Σ_(n=0) ^∞ (−x^2 )^n =(1/(1+x^2 )) ⇒ S(x) =∫_0 ^x (dt/(1+t^2 )) +c =arctan(x)+c c =S(o) =0 ⇒S(x) =arctanx and Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) =S(1) =arctan(1)=(π/4) .

$${let}\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:{due}\:{to}\:{uniform}\:{convergence}\:{we}\:{have} \\ $$$$\frac{{d}}{{dx}}{S}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{x}^{\mathrm{2}} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:+{c}\:={arctan}\left({x}\right)+{c}\: \\ $$$${c}\:={S}\left({o}\right)\:=\mathrm{0}\:\Rightarrow{S}\left({x}\right)\:={arctanx}\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:={S}\left(\mathrm{1}\right)\:={arctan}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

S_n =(1/1)−(1/3)+(1/5)−(1/7)+...∞ Gregory series tan^(−1) x=∫(dx/(1+x^2 )) tan^(−1) x=∫(1−x^2 +x^4 −x^6 +...)dx =x−(x^3 /3)+(x^5 /5)−(x^7 /7)+... put x=1 both side tan^(−1) (1)=1−(1/3)+(1/5)−(1/7)+... So S_n =tan^(−1) (1)=(π/4)

$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+…\infty \\ $$$${Gregory}\:{series} \\ $$$${tan}^{−\mathrm{1}} {x}=\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${tan}^{−\mathrm{1}} {x}=\int\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} −{x}^{\mathrm{6}} +…\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+… \\ $$$${put}\:{x}=\mathrm{1}\:{both}\:{side} \\ $$$${tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+… \\ $$$${So}\:{S}_{{n}} ={tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19