<s>[INST] Finding triple sum of the dependant variables method checking 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 ?

 [/INST] 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*}
</s>