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Question Number 29505 by abdo imad last updated on 09/Feb/18

let give u_n = Σ_(k=0) ^∞ (1/((k+1)^2 2^k )) find lim_(n→∞) u_n .

$${let}\:{give}\:{u}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{2}^{{k}} }\:\:{find}\:\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} \:\:. \\ $$

Commented by abdo imad last updated on 11/Feb/18

u_n =Σ_(k=0) ^n (1/((k+1)^2 2^k )).

$${u}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}^{{k}} }. \\ $$

Commented by abdo imad last updated on 13/Feb/18

we have u_n = Σ_(k=1) ^(n+1) (1/(k^2 2^(k−1) ))=2Σ_(k=1) ^(n+1) (1/k^2 )((1/2))^k but thserie Σ_(k=1) ^∞ (1/k^2 )((1/2))^k is convrgente let putw(x)=Σ_(n=1) ^∞ (x^n /n^2 ) for ∣x∣<1 we have w^′ (x)= Σ_(n=1) ^∞ (x^(n−1) /n) ⇒xw^′ (x)=Σ_(n=1) ^∞ (x^n /n) ⇒w^′ (x) +xw^(′′) (x)=Σ_(n=1) ^∞ x^(n−1) =Σ_(n=0) ^∞ x^n =(1/(1−x)) so w is solution for the d.e y^′ +xy^(′′) =(1/(1−x)) for y^′ =z ⇒ z+xz^′ =(1/(1−x)) h.e⇒z+xz^′ =0 ⇒xz^′ =−z ⇒(z^′ /z) =((−1)/x) ⇒ln∣z∣=−ln∣x∣ +k⇒z= (λ/x) for 0<x<1 mvc ⇒ z^′ =(λ^′ /x) −(λ/x^2 ) ⇒(λ/x) +λ^′ −(λ/x)=(1/(1−x)) ⇒λ^′ = (1/(1−x))⇒ λ(x)= −ln(1−x) +k and k=λ(0)=0 z(x)=−((ln(1−x))/x)=y^′ ⇒ y(x)=−∫_0 ^x ((ln(1−t))/t)dt +c c=y(0)=0 ⇒ w(x)= −∫_0 ^x ((ln(1−t))/t)dt and S=2w((1/2))=−2∫_0 ^(1/2) ((ln(1−t))/t)dt .....be continued....

$${we}\:{have}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \:\mathrm{2}^{{k}−\mathrm{1}} }=\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} {but}\:{thserie} \\ $$$$\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} \:{is}\:{convrgente}\:\:{let}\:{putw}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${for}\:\mid{x}\mid<\mathrm{1}\:\:{we}\:{have}\:{w}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{{n}}\:\Rightarrow{xw}^{'} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}} \\ $$$$\Rightarrow{w}^{'} \left({x}\right)\:+{xw}^{''} \left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}−\mathrm{1}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{so}\:{w}\:{is} \\ $$$${solution}\:{for}\:{the}\:{d}.{e}\:\:{y}^{'} \:+{xy}^{''} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{for}\:{y}^{'} ={z}\:\Rightarrow \\ $$$${z}+{xz}^{'} =\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:\:{h}.{e}\Rightarrow{z}+{xz}^{'} =\mathrm{0}\:\Rightarrow{xz}^{'} =−{z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{−\mathrm{1}}{{x}} \\ $$$$\Rightarrow{ln}\mid{z}\mid=−{ln}\mid{x}\mid\:+{k}\Rightarrow{z}=\:\frac{\lambda}{{x}}\:\:{for}\:\mathrm{0}<{x}<\mathrm{1}\:{mvc}\:\Rightarrow \\ $$$${z}^{'} =\frac{\lambda^{'} }{{x}}\:−\frac{\lambda}{{x}^{\mathrm{2}} }\:\Rightarrow\frac{\lambda}{{x}}\:+\lambda^{'} \:−\frac{\lambda}{{x}}=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow\lambda^{'} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\Rightarrow \\ $$$$\lambda\left({x}\right)=\:−{ln}\left(\mathrm{1}−{x}\right)\:+{k}\:{and}\:{k}=\lambda\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${z}\left({x}\right)=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}={y}^{'} \:\Rightarrow\:{y}\left({x}\right)=−\int_{\mathrm{0}} ^{{x}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}\:+{c} \\ $$$${c}={y}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{w}\left({x}\right)=\:−\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}\:{and} \\ $$$${S}=\mathrm{2}{w}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}\:.....{be}\:{continued}.... \\ $$

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

we have lim_(n→∞) u_n = 2Σ_(n=1) ^∞ (1/n^2 )((1/2))^n let put S(x)=Σ_(n=0) ^∞ x^n =(1/(1−x))⇒ ∫_0 ^x S(t)dt = Σ_(n=0) ^∞ (x^(n+1) /(n+1))=Σ_(n=1) ^∞ (x^n /n) =−ln(1−x) for 0<x<1 (λ=0)⇒ x Σ_(n=1) ^∞ (x^(n−1) /n)=−ln(1−x)⇒Σ_(n=1) ^∞ (x^(n−1) /n) =−((ln(1−x))/x)⇒ ∫_0 ^x (Σ_(n=1) ^∞ (t^(n−1) /n))dt=−∫_0 ^x ((ln(1−t))/t) +λ⇒ Σ_(n=1) ^∞ (x^n /n^2 ) =−∫_0 ^x ((ln(1−t))/t) (λ=0)and 2Σ_(n=1) ^∞ (1/n^2 )((1/2))^n =−∫_0 ^(1/2) ((ln(1−t))/t)dt ch. t=cosθ ∫_0 ^(1/2) (...)dt= ∫_(π/2) ^(π/3) ((ln(1−cosθ))/(cosθ))sinθ dθ =∫_(π/2) ^(π/3) tanθ ln(1−cosθ)dθ....be continued...

$${we}\:{have}\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} =\:\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \:{let}\:{put} \\ $$$${S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\Rightarrow\:\int_{\mathrm{0}} ^{{x}} {S}\left({t}\right){dt}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}} \\ $$$$=−{ln}\left(\mathrm{1}−{x}\right)\:\:{for}\:\mathrm{0}<{x}<\mathrm{1}\:\:\:\left(\lambda=\mathrm{0}\right)\Rightarrow \\ $$$${x}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}−\mathrm{1}} }{{n}}=−{ln}\left(\mathrm{1}−{x}\right)\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{{n}}\:=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}−\mathrm{1}} }{{n}}\right){dt}=−\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:+\lambda\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\:=−\int_{\mathrm{0}} ^{{x}} \:\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:\:\:\:\:\:\left(\lambda=\mathrm{0}\right){and} \\ $$$$\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \:=−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}\:\:{ch}.\:\:{t}={cos}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left(...\right){dt}=\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{ln}\left(\mathrm{1}−{cos}\theta\right)}{{cos}\theta}{sin}\theta\:{d}\theta \\ $$$$=\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{3}}} \:{tan}\theta\:{ln}\left(\mathrm{1}−{cos}\theta\right){d}\theta....{be}\:{continued}... \\ $$

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