Question

Solve the given expression.

$\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k}$

Given the condition that i ≠ j ≠ k.

Answer is 81/208 need solution.

Answer 1

To Evaluate :

$\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k}$

Solution :

Case 1 : i≠j≠k

$\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k}$

$\implies\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \left(\frac{1}{ {3}^{i} \cdot {3}^{j} } \right)\sum_{k=0}^\infty \frac{1}{3^k}$

${ \implies\displaystyle\sum_{i=0}^\infty \left( \frac{1}{ {3}^{i} } \right) \cdot \sum_{j=0}^\infty \left(\frac{1}{ {3}^{j} } \right) \cdot\sum_{k=0}^\infty \frac{1}{3^k}}$

$\displaystyle \rm \implies s_i \cdot s_j \cdot s_k$

Now, first we will calculate the value of (Si)

${\implies \displaystyle \rm s_i = \frac{1}{ {3}^{0} } + \frac{1}{3 {}^{1} } + \frac{1}{ {3}^{2} } + \frac{1}{ {3}^{3} } ... + \frac{1}{ {3}^{ \infin} } }$

Now, we can clearly see that a GP (geometric Progression) is formed.

So, The sum :

${\implies \displaystyle \rm s_i = \frac{a}{1 - r} }$

${\implies \displaystyle \rm s_i = \frac{1}{1 - \frac{1}{3} } }$

${\implies \displaystyle \rm s_i = \frac{1}{ \frac{2}{3} } }$

${\implies \displaystyle \rm s_i = \frac{3}{2} }$

$\implies\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k} = \frac{27}{8}$

Let this Be S1

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Case 2 : i=j=k

$\implies\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k}$

$\implies\displaystyle\sum_{i=0}^\infty \frac{1}{3^{3i}}$

$\implies\displaystyle\sum_{i=0}^\infty \frac{1}{3^{3i}} = \rm s_2$

$\implies\displaystyle \rm s_2 = \frac{1}{ {3}^{0} } + \frac{1 }{ {3}^{3} } + \frac{1}{ {3}^{6} } ... + \frac{1}{ {3}^{ \infin} }$

$\implies\displaystyle \rm s_2 = \frac{1}{ 1 - \frac{1}{3^{3} } }$

$\implies\displaystyle \rm s_2 = \frac{27}{ 26}$

$\implies\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k} = \frac{27}{26}$

let this be S2

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Case 3 : i=j≠k

$\implies\displaystyle\sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{1}{3^i3^j3^k}$

$\implies\displaystyle\sum_{i=0}^\infty \frac{1}{ {3}^{2i} } \sum_{k=0}^\infty \frac{1}{3^k}$

$\implies\displaystyle\sum_{i=0}^\infty \frac{1}{ {3}^{2i} } \left( \frac{3}{2} - \frac{1}{ {3}^{i} } \right)$

$\implies\displaystyle\sum_{i=0}^\infty \frac{1}{ {3}^{2i} } \cdot\frac{3}{2} - \frac{1}{ {3}^{3i} }$

$\implies\displaystyle \frac{3}{2} \sum_{i=0}^\infty \frac{1}{ {3}^{2i} } - \sum_{i=0}^\infty \frac{1}{3^{3i}}$

$\implies\displaystyle \frac{3}{2} \sum_{i=0}^\infty \frac{1}{ {3}^{2i} } - \sum_{i=0}^\infty \frac{1}{3^{3i}}$

$\implies\displaystyle \frac{3}{2} \cdot\frac{9}{ 8} - \frac{27}{26}$

$\implies\displaystyle \frac{3}{2} \cdot\frac{9}{ 8} - \frac{27}{26}$

$\implies\displaystyle \frac{135}{208}$

Let this Be S3

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So, the total sum will be :

S = S1-S2-3(S3)

(Because there can be 3 cases similar like S3)

$\implies \displaystyle \rm s = \frac{27}{8} - \frac{27}{26} - 3 \cdot \frac{135}{208}$

$\implies \displaystyle \rm s = \frac{81}{208}$

So, the solution is 81/208