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Question Number 154593 by Niiicooooo last updated on 19/Sep/21

Commented by Niiicooooo last updated on 19/Sep/21

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

S=Σ_(n=1) ^∞ ((n!)/((2n+1)!!))∙(1/((n+1)))=Σ_(n=1) ^∞ ((n!)/((2n+1)!))∙((2^n n!)/((n+1))) =Σ_(n=1) ^∞ (((n!)^2 )/((2n+1)!))∙(2^n /(n+1))=Σ_(n=1) ^∞ (2^n /(n+1))β(n+1, n+1) =∫_0 ^1 Σ_(n=1) ^∞ (((2x−2x^2 )^n )/(n+1))dx=−∫_0 ^1 (1+((ln(2x^2 −2x+1))/(2x^2 −2x)))dx Σ_(n=1) ^∞ t^n =(t/(1−t))=(1/(1−t))−1 ⇒Σ_(n=1) ^∞ (t^n /(n+1))=−1−((ln∣1−t∣)/t)

S=∑∞n=1n!(2n+1)!!⋅1(n+1)=∑∞n=1n!(2n+1)!⋅2nn!(n+1)=∑∞n=1(n!)2(2n+1)!⋅2nn+1=∑∞n=12nn+1β(n+1,n+1)=∫01∑∞n=1(2x−2x2)nn+1dx=−∫01(1+ln(2x2−2x+1)2x2−2x)dx∑∞n=1tn=t1−t=11−t−1⇒∑∞n=1tnn+1=−1−ln∣1−t∣t

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