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Question Number 46607 by maxmathsup by imad last updated on 29/Oct/18
calculate Σ_((i,j)∈ N^2 ) ((i+j)/3^(i+j) )
$${calculate}\:\sum_{\left({i},{j}\right)\in\:{N}^{\mathrm{2}} } \:\:\:\:\:\frac{{i}+{j}}{\mathrm{3}^{{i}+{j}} } \\ $$
Commented by maxmathsup by imad last updated on 30/Oct/18
we have S_n =Σ_(i=0) ^∞ ( Σ_(j=0) ^∞ ((i+j)/3^(i+j) )) =Σ_(i=0) ^∞ S_i S_i =(i/3^i )Σ_(j=0) ^∞ (1/3^j ) +(1/3^i ) Σ_(j=0) ^∞ (j/3^j ) but Σ_(j=0) ^∞ (1/3^j ) =(1/(1−(1/3))) =(3/2) also Σ_(j=0) ^∞ (j/3^j ) =Σ_(j=1) ^∞ j((1/3))^j =w((1/3)) with w(x)=Σ_(n=0) ^∞ nx^n we have Σ_(n=0) ^∞ x^n =(1/(1−x)) for ∣x∣<1 ⇒Σ_(n=1) ^∞ nx^(n−1) =(x/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ nx^n =(x^2 /((1−x)^2 )) ⇒w((1/3))=(1/(9((2/3))^2 )) =(1/4) ⇒S_i =(3/2)(i/3^i ) +(1/(4.3^i )) ⇒ Σ_(i=0) ^∞ S_i =(3/2) Σ_(i=0) ^∞ (i/3^i ) +(1/4) Σ_(i=0) ^∞ (1/3^i ) =(3/2) (1/4) +(1/4) (3/2) =(3/8) +(3/8) =(6/8) =(3/4) ⇒ S=(3/4) .
$${we}\:{have}\:{S}_{{n}} =\sum_{{i}=\mathrm{0}} ^{\infty} \left(\:\sum_{{j}=\mathrm{0}} ^{\infty} \:\frac{{i}+{j}}{\mathrm{3}^{{i}+{j}} }\right)\:=\sum_{{i}=\mathrm{0}} ^{\infty} \:{S}_{{i}} \\ $$$${S}_{{i}} =\frac{{i}}{\mathrm{3}^{{i}} }\sum_{{j}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{3}^{{j}} }\:+\frac{\mathrm{1}}{\mathrm{3}^{{i}} }\:\sum_{{j}=\mathrm{0}} ^{\infty} \:\frac{{j}}{\mathrm{3}^{{j}} }\:{but}\:\sum_{{j}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{3}^{{j}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:{also} \\ $$$$\sum_{{j}=\mathrm{0}} ^{\infty} \:\frac{{j}}{\mathrm{3}^{{j}} }\:=\sum_{{j}=\mathrm{1}} ^{\infty} \:{j}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{j}} \:={w}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:{with}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{nx}^{{n}} \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:{for}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}−\mathrm{1}} =\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:=\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow{w}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{9}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{S}_{{i}} =\frac{\mathrm{3}}{\mathrm{2}}\frac{{i}}{\mathrm{3}^{{i}} }\:+\frac{\mathrm{1}}{\mathrm{4}.\mathrm{3}^{{i}} }\:\Rightarrow \\ $$$$\sum_{{i}=\mathrm{0}} ^{\infty} \:{S}_{{i}} =\frac{\mathrm{3}}{\mathrm{2}}\:\sum_{{i}=\mathrm{0}} ^{\infty} \:\frac{{i}}{\mathrm{3}^{{i}} }\:+\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{i}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{3}^{{i}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\mathrm{3}}{\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{8}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:=\frac{\mathrm{6}}{\mathrm{8}}\:=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow\:{S}=\frac{\mathrm{3}}{\mathrm{4}}\:. \\ $$$$ \\ $$

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