Question Number 59438 by tanmay last updated on 10/May/19
Commented by maxmathsup by imad last updated on 10/May/19
![we have I_n =_(πx =t) ∫_0 ^π (t^n /π^n ) sin(t) (dt/π) =(1/π^(n+1) ) ∫_0 ^π t^n sint dt let find W_n =∫_0 ^π t^n sint dt by parts u^′ =t^n and v=sint ⇒ W_n =[(1/(n+1))t^(n+1) sint]_0 ^π −∫_0 ^π (t^(n+1) /(n+1)) cost dt =−(1/(n+1)){ [(1/(n+2))t^(n+2) cost]_0 ^π +∫_0 ^π (t^(n+2) /(n+2))sint} =−(1/(n+1)){(1/(n+2))(−π^(n+2) ) +(1/(n+2)) ∫_0 ^π t^(n+2) sint dt} =(π^(n+2) /((n+1)(n+2))) −(1/((n+1)(n+2))) ∫_0 ^π t^(n+2) sint dt ⇒W_n =(π^(n+2) /((n+1)(n+2))) −(1/((n+1)(n+2))) W_(n+2) we have I_n =(1/π^(n+1) ) W_n ⇒I_n =(π/((n+1)(n+2))) −(1/(π^(n+1) (n+1)(n+2))) W_(n+2) =(π/((n+1)(n+2))) −(1/(π^(n+1) (n+1)(n+2))) π^(n+3) I_(n+2) ⇒ I_n =(π/((n+1)(n+2))) −(π^2 /((n+1)(n+2))) I_(n+2) =(π/((n+1)(n+2))){1−π I_(n+2) } ⇒ (I_n /(1−π I_(n+2) )) =(π/((n+1)(n+2))) ⇒Σ_(n=0) ^∞ (I_n /(1−πI_(n+2) )) =π Σ_(n=0) ^∞ (1/((n+1)(n+2))) =π Σ_(n=1) ^∞ (1/(n(n+1))) let S_n =Σ_(k=1) ^n (1/(k(k+1))) ⇒S_n =Σ_(k=1) ^n ((1/k) −(1/(k+1))) =1−(1/2) +(1/2) −(1/3) +....(1/n) −(1/(n+1)) =1−(1/(n+1)) →1 (n→+∞) ⇒ Σ_(n=0) ^∞ { (I_n /(1−π I_(n+2) ))} =π .](https://www.tinkutara.com/question/Q59479.png)
$${we}\:{have}\:{I}_{{n}} =_{\pi{x}\:={t}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{t}^{{n}} }{\pi^{{n}} }\:{sin}\left({t}\right)\:\frac{{dt}}{\pi}\:=\frac{\mathrm{1}}{\pi^{{n}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\pi} \:\:{t}^{{n}} {sint}\:{dt}\:\:{let}\:{find}\: \\ $$$${W}_{{n}} =\int_{\mathrm{0}} ^{\pi} \:{t}^{{n}} {sint}\:{dt}\:\:\:{by}\:{parts}\:\:{u}^{'} ={t}^{{n}} \:{and}\:{v}={sint}\:\Rightarrow \\ $$$${W}_{{n}} =\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} {sint}\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:{cost}\:{dt}\:=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\:\:\left[\frac{\mathrm{1}}{{n}+\mathrm{2}}{t}^{{n}+\mathrm{2}} {cost}\right]_{\mathrm{0}} ^{\pi} +\int_{\mathrm{0}} ^{\pi} \frac{{t}^{{n}+\mathrm{2}} }{{n}+\mathrm{2}}{sint}\right\} \\ $$$$=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\frac{\mathrm{1}}{{n}+\mathrm{2}}\left(−\pi^{{n}+\mathrm{2}} \right)\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:{t}^{{n}+\mathrm{2}} \:{sint}\:{dt}\right\} \\ $$$$=\frac{\pi^{{n}+\mathrm{2}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:\int_{\mathrm{0}} ^{\pi} \:\:{t}^{{n}+\mathrm{2}} {sint}\:{dt}\:\Rightarrow{W}_{{n}} =\frac{\pi^{{n}+\mathrm{2}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:{W}_{{n}+\mathrm{2}} \\ $$$${we}\:{have}\:{I}_{{n}} =\frac{\mathrm{1}}{\pi^{{n}+\mathrm{1}} }\:{W}_{{n}} \:\Rightarrow{I}_{{n}} =\frac{\pi}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\pi^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:{W}_{{n}+\mathrm{2}} \\ $$$$=\frac{\pi}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\pi^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:\pi^{{n}+\mathrm{3}} \:{I}_{{n}+\mathrm{2}} \:\Rightarrow \\ $$$${I}_{{n}} =\frac{\pi}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:−\frac{\pi^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:{I}_{{n}+\mathrm{2}} =\frac{\pi}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\left\{\mathrm{1}−\pi\:{I}_{{n}+\mathrm{2}} \right\}\:\Rightarrow \\ $$$$\frac{{I}_{{n}} }{\mathrm{1}−\pi\:{I}_{{n}+\mathrm{2}} }\:=\frac{\pi}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{I}_{{n}} }{\mathrm{1}−\pi{I}_{{n}+\mathrm{2}} }\:=\pi\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\pi\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:\:{let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\:\:\:\Rightarrow{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(\frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{3}}\:+….\frac{\mathrm{1}}{{n}}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:=\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\rightarrow\mathrm{1}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \left\{\:\:\frac{{I}_{{n}} }{\mathrm{1}−\pi\:{I}_{{n}+\mathrm{2}} }\right\}\:=\pi\:. \\ $$
Commented by tanmay last updated on 10/May/19
$${excellent}\:{sir}… \\ $$