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