Σ_(n=0) ^∞ ((2n+1)/2^(2n+1) )=?


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Question Number 3807 by Rasheed Soomro last updated on 21/Dec/15

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }=? \\ $$
	Answered by Yozzii last updated on 21/Dec/15
	$s=Σ_(n=0) ^∞ ((2n+1)/2^(2n+1) )=(1/2)Σ_(n=0) ^∞ ((2n+1)/((4)^n )) s=(1/2){2Σ_(n=0) ^∞ (n/4^n )+Σ_(n=0) ^∞ (1/4^n )} s=Σ_(n=0) ^∞ n(0.25)^n +(1/2)Σ_(n=0) ^∞ (1/4^n ) Let l=Σ_(n=0) ^∞ (n/4^n )=0+Σ_(n=1) ^∞ (n/4^n )=(1/4)+(2/4^2 )+(3/4^3 )+(4/4^4 )+(5/4^5 )... 4l=1+(2/4)+(3/4^2 )+(4/4^3 )+(5/4^4 )+... 4l−l=1+(1/4)+(1/4^2 )+(1/4^3 )+(1/4^4 )+... 3l=1+((1/4)/(1−1/4)) 3l=1+((1/4)/(3/4))=(4/3) l=(4/9) s=(4/9)+(1/2)×(1/(1−(1/4)))=(4/9)+(1/2)×(4/3) s=(4/9)+(2/3)=((12+18)/(27))=((4+6)/9)=((10)/9)$
	$${s}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{4}\right)^{{n}} } \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{4}^{{n}} }+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}} }\right\} \\ $$$${s}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{n}\left(\mathrm{0}.\mathrm{25}\right)^{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}} } \\ $$$${Let}\:{l}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{4}^{{n}} }=\mathrm{0}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{4}^{{n}} }=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{4}^{\mathrm{4}} }+\frac{\mathrm{5}}{\mathrm{4}^{\mathrm{5}} }... \\ $$$$\mathrm{4}{l}=\mathrm{1}+\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{4}^{\mathrm{4}} }+... \\ $$$$\mathrm{4}{l}−{l}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }+... \\ $$$$\mathrm{3}{l}=\mathrm{1}+\frac{\mathrm{1}/\mathrm{4}}{\mathrm{1}−\mathrm{1}/\mathrm{4}} \\ $$$$\mathrm{3}{l}=\mathrm{1}+\frac{\mathrm{1}/\mathrm{4}}{\mathrm{3}/\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${l}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${s}=\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${s}=\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{12}+\mathrm{18}}{\mathrm{27}}=\frac{\mathrm{4}+\mathrm{6}}{\mathrm{9}}=\frac{\mathrm{10}}{\mathrm{9}} \\ $$$$ \\ $$$$ \\ $$
	Commented by Rasheed Soomro last updated on 22/Dec/15

	$$\overset{\mathbb{V}} {\mathbb{G}_{\mathbb{OO}} \:\mathbb{D}}\:\:\:{for}\:{your}\:\mathbb{G}\mathrm{rip}\:{on}\:{maths}! \\ $$
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