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Question Number 31743 by rahul 19 last updated on 13/Mar/18

Find the value of : Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/(3^i 3^j 3^k )). case 1: i≠j≠k. case 2: i<j<k.

$$\:{Find}\:{the}\:{value}\:{of}\::\:\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} \mathrm{3}^{{k}} }. \\ $$ $${case}\:\mathrm{1}:\:{i}\neq{j}\neq{k}. \\ $$ $${case}\:\mathrm{2}:\:{i}<{j}<{k}. \\ $$

Answered by mrW2 last updated on 13/Mar/18

Σ_(k=0) ^∞ (1/3^k )=1+(1/3)+(1/3^2 )+....=(1/(1−(1/3)))=(3/2) Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/(3^i 3^j 3^k )) =Σ_(i=0) ^∞ Σ_(j=0) ^∞ {(1/(3^i 3^j ))(Σ_(k=0) ^∞ (1/3^k ))} =(3/2)Σ_(i=0) ^∞ Σ_(j=0) ^∞ (1/(3^i 3^j )) =(3/2)Σ_(i=0) ^∞ {(1/3^i )(Σ_(j=0) ^∞ (1/3^j ))} =(3/2)×(3/2)Σ_(i=0) ^∞ (1/3^i ) =(3/2)×(3/2)×(3/2) =((27)/8)

$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{k}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+....=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} \mathrm{3}^{{k}} } \\ $$ $$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} }\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{{k}} }\right)\right\} \\ $$ $$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} } \\ $$ $$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{3}^{{i}} }\left(\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{j}} }\right)\right\} \\ $$ $$=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{i}} } \\ $$ $$=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$=\frac{\mathrm{27}}{\mathrm{8}} \\ $$

Commented bymrW2 last updated on 14/Mar/18

Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/(3^i 3^j 3^k ))=((27)/8) this is correct. if you need Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ _(i≠j≠k) (1/(3^i 3^j 3^k )) or Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ _(i>j>k) (1/(3^i 3^j 3^k )) then the result is different.

$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} \mathrm{3}^{{k}} }=\frac{\mathrm{27}}{\mathrm{8}} \\ $$ $${this}\:{is}\:{correct}. \\ $$ $${if}\:{you}\:{need}\:\underset{{i}\neq{j}\neq{k}} {\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}}\:\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} \mathrm{3}^{{k}} }\:{or} \\ $$ $$\underset{{i}>{j}>{k}} {\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}}\:\frac{\mathrm{1}}{\mathrm{3}^{{i}} \mathrm{3}^{{j}} \mathrm{3}^{{k}} }\:{then}\:{the}\:{result}\:{is}\:{different}. \\ $$

Commented byrahul 19 last updated on 14/Mar/18

But what is the difference when i≠j≠k and i<j<k are present in ques. (original q.) and when that is not mention in the ques. (the way u solved).

$${But}\:{what}\:{is}\:{the}\:{difference}\:{when} \\ $$ $${i}\neq{j}\neq{k}\:{and}\:{i}<{j}<{k}\:{are}\:{present}\:{in}\:{ques}. \\ $$ $$\left({original}\:{q}.\right)\:{and}\:{when}\:{that}\:{is}\:{not}\:{mention} \\ $$ $${in}\:{the}\:{ques}.\:\left({the}\:{way}\:{u}\:{solved}\right). \\ $$

Commented bymrW2 last updated on 14/Mar/18

look at an example: Σ_(i=1) ^3 Σ_(j=1) ^3 a^i b^j =a^1 (b^1 +b^2 +b^3 )+a^2 (b^1 +b^2 +b^3 )+a^3 (b^1 +b^2 +b^3 )=(a^1 +a^2 +a^3 )(b^1 +b^2 +b^3 ) (i and j are independent) Σ_(i=1) ^3 Σ_(j=1) ^3 _(i≠j) a^i b^j =a^1 (b^2 +b^3 )+a^2 (b^1 +b^3 )+a^3 (b^1 +b^2 ) (i and j are dependent) Σ_(i=1) ^3 Σ_(j=1) ^3 _(i<j) a^i b^j =a^1 (b^2 +b^3 )+a^2 b^3 (i and j are dependent)

$${look}\:{at}\:{an}\:{example}: \\ $$ $$\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{a}^{{i}} {b}^{{j}} ={a}^{\mathrm{1}} \left({b}^{\mathrm{1}} +{b}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)+{a}^{\mathrm{2}} \left({b}^{\mathrm{1}} +{b}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)+{a}^{\mathrm{3}} \left({b}^{\mathrm{1}} +{b}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)=\left({a}^{\mathrm{1}} +{a}^{\mathrm{2}} +{a}^{\mathrm{3}} \right)\left({b}^{\mathrm{1}} +{b}^{\mathrm{2}} +{b}^{\mathrm{3}} \right) \\ $$ $$\left({i}\:{and}\:{j}\:{are}\:{independent}\right) \\ $$ $$ \\ $$ $$\underset{{i}\neq{j}} {\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}}{a}^{{i}} {b}^{{j}} ={a}^{\mathrm{1}} \left({b}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)+{a}^{\mathrm{2}} \left({b}^{\mathrm{1}} +{b}^{\mathrm{3}} \right)+{a}^{\mathrm{3}} \left({b}^{\mathrm{1}} +{b}^{\mathrm{2}} \right) \\ $$ $$\left({i}\:{and}\:{j}\:{are}\:{dependent}\right) \\ $$ $$ \\ $$ $$\underset{{i}<{j}} {\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}}{a}^{{i}} {b}^{{j}} ={a}^{\mathrm{1}} \left({b}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)+{a}^{\mathrm{2}} {b}^{\mathrm{3}} \\ $$ $$\left({i}\:{and}\:{j}\:{are}\:{dependent}\right) \\ $$

Commented byrahul 19 last updated on 14/Mar/18

ok sir, now understood when its in 2 variables (i,j). pls also show similar example for 3 variable (i,j,k) ?

$${ok}\:{sir},\:{now}\:{understood}\:{when}\:{its}\:{in} \\ $$ $$\mathrm{2}\:{variables}\:\left({i},{j}\right). \\ $$ $${pls}\:{also}\:{show}\:{similar}\:{example}\:{for} \\ $$ $$\mathrm{3}\:{variable}\:\left({i},{j},{k}\right)\:? \\ $$

Commented byabdo imad last updated on 14/Mar/18

there is a dfference between Σ_(i j) a_i bj and Σ_(i,j andi≠j) a_i a_j and the type of this triple sum is difficut to treat and if you give interest you must take a look to sommables family ....

$${there}\:{is}\:{a}\:{dfference}\:{between}\:\:\sum_{{i}\:{j}} {a}_{{i}} {bj}\:{and}\:\:\sum_{{i},{j}\:{andi}\neq{j}} {a}_{{i}} {a}_{{j}} \\ $$ $${and}\:{the}\:{type}\:{of}\:{this}\:{triple}\:{sum}\:{is}\:{difficut}\:{to}\:{treat} \\ $$ $${and}\:{if}\:{you}\:{give}\:{interest}\:{you}\:{must}\:{take}\:{a}\:{look}\:{to} \\ $$ $${sommables}\:{family}\:.... \\ $$

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