find-n-0-n-1-4-n-

Question Number 29986 by abdo imad last updated on 14/Feb/18

$${find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{n}+\mathrm{1}}{\mathrm{4}^{{n}} }\:. \\ $$

Commented by abdo imad last updated on 14/Feb/18

let introduce for ∣x∣<1 S(x)= Σ_(n=0) ^∞ (n+1)x^n we have Σ_(n=0) ^∞ ((n+1)/4^n ) =S((1/4)) we have ∫_0 ^x S(t)dt =Σ_(n=0) ^∞ x^(n+1) +λ x=0 ⇒λ=0 and ∫_0 ^x S(t)dt=Σ_(n=1) ^∞ x^n =(1/(1−x)) −1 =(x/(1−x)) ⇒ S(x)=(d/dx)( (x/(1−x)))= ((1−x −x(−1))/((1 −x)^2 ))= (1/((1−x)^2 )) ⇒ S((1/4))= (1/(((3/4))^2 )) = ((16)/9) ⇒ Σ_(n=0) ^∞ ((n+1)/4^n ) = ((16)/9) .

$${let}\:{introduce}\:{for}\:\mid{x}\mid<\mathrm{1}\:\:\:{S}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left({n}+\mathrm{1}\right){x}^{{n}} \:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}+\mathrm{1}}{\mathrm{4}^{{n}} }\:={S}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\:{we}\:{have}\:\:\int_{\mathrm{0}} ^{{x}} \:{S}\left({t}\right){dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}+\mathrm{1}} \:\:+\lambda \\ $$$${x}=\mathrm{0}\:\Rightarrow\lambda=\mathrm{0}\:{and}\:\int_{\mathrm{0}} ^{{x}} \:{S}\left({t}\right){dt}=\sum_{{n}=\mathrm{1}} ^{\infty} {x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:−\mathrm{1}\:=\frac{{x}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\:{S}\left({x}\right)=\frac{{d}}{{dx}}\left(\:\frac{{x}}{\mathrm{1}−{x}}\right)=\:\frac{\mathrm{1}−{x}\:−{x}\left(−\mathrm{1}\right)}{\left(\mathrm{1}\:−{x}\right)^{\mathrm{2}} }=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\:\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{16}}{\mathrm{9}}\:\Rightarrow\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{n}+\mathrm{1}}{\mathrm{4}^{{n}} }\:=\:\frac{\mathrm{16}}{\mathrm{9}}\:. \\ $$

Answered by MJS last updated on 14/Feb/18

Σ_(n=0) ^∞ ((n+1)/k^n )=(k^2 /((k−1)^2 )) with k>1

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}+\mathrm{1}}{{k}^{{n}} }=\frac{{k}^{\mathrm{2}} }{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{with}\:{k}>\mathrm{1} \\ $$

Commented by abdo imad last updated on 14/Feb/18

$${you}\:{must}\:{show}\:{your}\:{work}\:{and}\:{your}\:{method}\:{sir}\:{mjs}… \\ $$

Commented by abdo imad last updated on 15/Feb/18

we have proved that Σ_(n=0) ^∞ (n+1)x^n = (1/((1−x)^2 )) for ∣x∣<1 and for x= (1/k) with k>1 we obtain Σ_(n=0) ^∞ ((n+1)/k^n ) = (1/((1−(1/k))^2 ))= (k^2 /((k−1)^2 )) .

$${we}\:{have}\:{proved}\:{that}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({n}+\mathrm{1}\right){x}^{{n}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$${and}\:{for}\:{x}=\:\frac{\mathrm{1}}{{k}}\:{with}\:{k}>\mathrm{1}\:{we}\:{obtain} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}+\mathrm{1}}{{k}^{{n}} }\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{{k}}\right)^{\mathrm{2}} }=\:\frac{{k}^{\mathrm{2}} }{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$

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