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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 .

wehaveIn=πx=t0πtnπnsin(t)dtπ=1πn+10πtnsintdtletfindWn=0πtnsintdtbypartsu=tnandv=sintWn=[1n+1tn+1sint]0π0πtn+1n+1costdt=1n+1{[1n+2tn+2cost]0π+0πtn+2n+2sint}=1n+1{1n+2(πn+2)+1n+20πtn+2sintdt}=πn+2(n+1)(n+2)1(n+1)(n+2)0πtn+2sintdtWn=πn+2(n+1)(n+2)1(n+1)(n+2)Wn+2wehaveIn=1πn+1WnIn=π(n+1)(n+2)1πn+1(n+1)(n+2)Wn+2=π(n+1)(n+2)1πn+1(n+1)(n+2)πn+3In+2In=π(n+1)(n+2)π2(n+1)(n+2)In+2=π(n+1)(n+2){1πIn+2}In1πIn+2=π(n+1)(n+2)n=0In1πIn+2=πn=01(n+1)(n+2)=πn=11n(n+1)letSn=k=1n1k(k+1)Sn=k=1n(1k1k+1)=112+1213+....1n1n+1=11n+11(n+)n=0{In1πIn+2}=π.

Commented by tanmay last updated on 10/May/19

excellent sir...

excellentsir...

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