Question-154593

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)

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}\centerdot\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\centerdot\frac{\mathrm{2}^{{n}} {n}!}{\left({n}+\mathrm{1}\right)} \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\centerdot\frac{\mathrm{2}^{{n}} }{{n}+\mathrm{1}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{n}} }{{n}+\mathrm{1}}\beta\left({n}+\mathrm{1},\:{n}+\mathrm{1}\right) \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} \right)^{{n}} }{{n}+\mathrm{1}}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{ln}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right){dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{t}^{{n}} =\frac{{t}}{\mathrm{1}−{t}}=\frac{\mathrm{1}}{\mathrm{1}−{t}}−\mathrm{1}\:\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{{n}} }{{n}+\mathrm{1}}=−\mathrm{1}−\frac{\mathrm{ln}\mid\mathrm{1}−{t}\mid}{{t}} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply