Question Number 100468 by Mikael_786 last updated on 26/Jun/20

Σ_(n=1) ^∞ (n/((2n+1)!)) help me pls

Answered by mathmax by abdo last updated on 26/Jun/20

S =(1/2)Σ_(n=0) ^∞ ((2n+1−1)/((2n+1)!)) =(1/2)Σ_(n=0) ^∞ (1/((2n)!))−(1/2)Σ_(n=0) ^∞ (1/((2n+1)!)) we have e^x +e^(−x) =Σ_(n=0) ^∞ (x^n /(n!)) +Σ_(n=0) ^∞ (((−1)^n x^n )/(n!)) =Σ_(n=0) ^∞ (1/(n!))(1+(−1)^n )x^n =2Σ_(n=0) ^∞ (x^(2n) /((2n)!)) ⇒((e^x +e^(−x) )/2) =Σ_(n=0) ^∞ (x^(2n) /((2n)!)) ⇒Σ_(n=0) ^∞ (1/((2n)!)) =((e+e^(−1) )/2) e^x −e^(−x) =Σ_(n=0) ^∞ (1/(n!))(1−(−1)^n )x^n =2 Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) ⇒ Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) =((e^x −e^(−x) )/2) ⇒Σ_(n=0) ^∞ (1/((2n+1)!)) =((e−e^(−1) )/2) ⇒ S =(1/4)(e+e^(−1) )−(1/4)(e−e^(−1) ) =(1/4)(e+e^(−1) −e+e^(−1) ) =((2e^(−1) )/4) =(1/(2e)) S =(1/(2e))

Commented byMikael_786 last updated on 27/Jun/20

thank you Sir

Answered by smridha last updated on 26/Jun/20

ans:(1/(2e)).apply generalised hypergeometric f^n

Answered by maths mind last updated on 26/Jun/20

f(x)=Σ_(k≥1) (x^(2n+1) /((2n+1)!))=((e^x −e^(−x) )/2) ⇒((e^x −e^(−x) )/(2x))=Σ_(n≥1) (x^(2n) /((2n+1)!)) ∂x(1/2) (((e^x −e^(−x) )/x))∣_(x=1) =Σ_(k≥1) ((2n)/((2n+1)!)) ⇔(1/z)(((e^x +e^(−x) )/x)−((e^x −e^(−x) )/x^2 ))∣_(x=1) =2Σ_(n=1) ^(+∞) (n/((2n+1)!)) ⇔e^(−1) =2Σ(n/((2n+1)!))⇒(1/(2e))=Σ_(n≥1) (n/((2n+1)!))

Commented byMikael_786 last updated on 27/Jun/20

thank you Sir