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Find:
A) $\sum_{i \geq j \leq k} \frac{1}{3^i 4^j 5^k}$ $i,j,k$ vary from $[0,\infty)$
B) $\sum_{i<j<k} \frac{1}{3^i 3^j 3^k}$ $i,j,k$ vary from $[0,\infty)$

My method as both are dependent summation , but second one is symmetric with the variables (function wise $\frac{1}{3^x}$ only ) so i thought maybe solving B gives hint for A) :for B) i can use the method as we do for two variable case , we know the product of those terms if they(variables) were all independent would be having these order of subscripts in the terms : $i=j=k , i<j>k,i<j<k,i<j=k,i>j>k,i>j=k,i>j<k ,i=j>k, i=j<k$ we can argue that product of the terms of forms $i<j<k$ would be same as $i>j>k$ as symmetric . Similarily for $i>j =k$ and $i=j<k$ and $i=j>k$ , $i<j=k$ So we get total sum for independent one to be equal to these individual order terms : $2(i=j<k) + 2(i=j>k) + (i=j=k) + 2(i<j<k) + (i<j>k) + (i>j<k)$ , we wanted the fourth sum in the above for B part , but for that i need to evaluate others how would i do so ? Or there is a different approach to one can think of too ?

Paracetamol
  • 624

1 Answers1

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Both triple sums can be treated in the same way by transforming the index regions to obtain a geometric series in the form $\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}$ valid for $|q|<1$.

Case A: We obtain \begin{align*} \color{blue}{\sum_{0\leq j\leq i,k<\infty}\frac{1}{3^i 4^j 5^k}} &=\sum_{j=0}^\infty\frac{1}{4^j}\left(\sum_{i=j}^\infty \frac{1}{3^i}\right)\left(\sum_{k=j}^\infty \frac{1}{5^k}\right)\tag{1}\\ &=\sum_{j=0}^\infty\frac{1}{4^j}\left(\sum_{i=0}^\infty \frac{1}{3^{i+j}}\right)\left(\sum_{k=0}^\infty \frac{1}{5^{k+j} }\right)\tag{2}\\ &=\sum_{j=0}^\infty\frac{1}{4^j3^j5^j}\left(\sum_{i=0}^\infty \frac{1}{3^{i}}\right)\left(\sum_{k=0}^\infty \frac{1}{5^{k} }\right)\tag{3}\\ &=\frac{1}{1-\frac{1}{60}}\,\frac{1}{1-\frac{1}{3}}\,\frac{1}{1-\frac{1}{5}}\tag{4}\\ &=\frac{60}{59}\cdot\frac{3}{2}\cdot\frac{5}{4}\\ &\,\,\color{blue}{=\frac{225}{118}\doteq1.906} \end{align*}

Comment:

  • In (1) we write the triple sum somewhat more conveniently by rearranging terms as preparation for the next steps.

  • In (2) shift the index of the inner sums to start with index $i=0$ and $k=0$.

  • In (3) we collect factors with like exponents.

  • In (4) we apply the geometric series expansion

Case B: We proceed similarly as in (A) and obtain \begin{align*} \color{blue}{\sum_{0\leq i<j<k<\infty}\frac{1}{3^i 3^j 3^k}} &=\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=i+1}^\infty\frac{1}{3^j}\sum_{k=j+1}^\infty\frac{1}{3^k}\\ &=\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=i+1}^\infty\frac{1}{3^j}\sum_{k=0}^\infty\frac{1}{3^{k+j+1}}\\ &=\frac{1}{3}\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=i+1}^\infty\frac{1}{9^j}\sum_{k=0}^\infty\frac{1}{3^{k}}\\ &=\frac{1}{3}\sum_{i=0}^\infty\frac{1}{3^i}\sum_{j=0}^\infty\frac{1}{9^{j+i+1}}\sum_{k=0}^\infty\frac{1}{3^{k}}\\ &=\frac{1}{27}\sum_{i=0}^\infty\frac{1}{27^i}\sum_{j=0}^\infty\frac{1}{9^{j}}\sum_{k=0}^\infty\frac{1}{3^{k}}\\ &=\frac{1}{27}\,\frac{1}{1-\frac{1}{27}}\,\frac{1}{1-\frac{1}{9}}\,\frac{1}{1-\frac{1}{3}}\\ &=\frac{1}{27}\cdot\frac{27}{26}\cdot\frac{9}{8}\cdot\frac{3}{2}\\ &\,\,\color{blue}{=\frac{27}{416}\doteq 0.064} \end{align*}

Markus Scheuer
  • 108,315