How to read and execute $\sum_{1 \leq \ell 
How to read and execute this sum?
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$$
I am having trouble to understand where is my error.
The question does not say, but I am assuming that $\ell$ starts at $1$, $m$ at $2$, and so $n$ at $3$.
This is essential a product of pg:
$$\sum_{1 \leq \ell<m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}\frac{1}{5^{\ell}}\sum_{m=2}\frac{1}{3^{m}}\sum_{n=3}\frac{1}{2^{n}} = \frac{1/5}{1-1/5}\frac{1/9}{1-1/3}\frac{1/8}{1-1/2}$$
But this does not agree with the answer :/
 A: *

*Assuming $n$ is not fixed:
Rewrite as triple sum
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}^\infty\sum_{m=\ell+1}^\infty\sum_{n=m+1}^\infty\frac{1}{5^{\ell}3^{m}2^{n}} =\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} $$
Then, this should be obvious in terms of how to read and execute.
Indeed, since each of them are Geometric Progressions, we have
$$\begin{align}\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} &= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \left(\frac{1}{2^m}\right) \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{6^m} \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell}\left(\frac{1}{5\cdot 6^\ell}\right) \\& = \frac{1}{5}\sum_{\ell=1}^\infty \frac{1}{30^\ell} \\&= \frac{1}{5}\cdot\frac{1}{29} \\&= \frac{1}{145}\end{align} $$



*Assuming $n$ is fixed:
The sum will be a finite double sum and you can exclude $1/2^n$ from the sum(treating that as a constant). Then, again use the GP formula to calculate each $\ell$-sum and $m$-sum.
A: The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq \ell <m<n}}\frac{1}{5^{\ell}3^m2^n}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq \ell <m<n
\end{align*}
which has $1$ as lower limit and $n$ as upper limit. We have two indices $\ell$ and $m$, which means we can write it as double sum as shown in the evaluation below.

We obtain
\begin{align*}
\color{blue}{\sum_{1\leq \ell <m<n}\frac{1}{5^{\ell}3^m2^n}}
&=\frac{1}{2^n}\sum_{\ell = 1}^{n-2}\frac{1}{5^{\ell}}\sum_{m=\ell+1}^{n-1}\frac{1}{3^m}\tag{2}\\
&=\frac{1}{2^n}\sum_{\ell = 1}^{n-2}\frac{1}{5^{\ell}}\left(\frac{\left(\frac{1}{3}\right)^{l+1}-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}\right)\tag{3}\\
&=\frac{1}{2^n}\sum_{\ell = 1}^{n-2}\frac{1}{5^l}\,\frac{1}{2}\left(\frac{1}{3^l}-\frac{1}{3^{n-1}}\right)\tag{4}\\
&= \frac{1}{2^{n+1}}\sum_{l=1}^{n-2}\frac{1}{15^l}- \frac{1}{2^{n+1}\,3^{n-1}}\sum_{l=1}^{n-2}\frac{1}{5^l}\\
&=\frac{1}{2^{n+1}}\left(\frac{\frac{1}{15}-\left(\frac{1}{15}\right)^{n-1}}{1-\frac{1}{15}}\right)-\frac{1}{2^{n+1}\,3^{n-1}}\left(\frac{\frac{1}{5}-\left(\frac{1}{5}\right)^{n-1}}{1-\frac{1}{5}}\right)\tag{5}\\
&=\frac{1}{2^{n+2}\cdot7}\left(1-\frac{1}{15^{n-2}}\right)-\frac{1}{2^{n+3 }\,3^{n-1}}\left(1-\frac{1}{5^{n-2}}\right)\\
&\,\,\color{blue}{=\frac{1}{2^{n+2}\cdot7}-\frac{1}{2^{n+3}\,3^{n-1}}+\frac{1}{2^{n+3}\,3^{n-1}\,5^{n-2}\cdot7}}
\end{align*}

Comment:

*

*In (2) we factor out $\frac{1}{2^n}$ and reorder the double sum using another common style.


*In (3) we evaluate the inner sum using the finite geometric summation formula.


*In (4) we do a simplification and multiply out in the next line.


*In (5) we apply the finite geometric summation formula twice and do a simplification in the following lines.

Note: A varying upper limit $n$ is not admissible in your case. Here $n$ is a free variable whereas the indices $\ell$ and $m$ are bound variables. This is different to the situation
\begin{align*}
\sum_{1\leq \ell <m<n\color{blue}{<\infty}}\frac{1}{5^{\ell}3^m2^n}
\end{align*}
where $n$ is bound by the upper limit $\infty$ and where $n$ varies between $m$ and $\infty$.
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
A: We will assume that $n$ is not fixed.
Firstly, note that for any triple $(l,m,n)\in\mathbb{N}^3$ such that $l<m<n$ there is exactly one triple $(a,b,c)\in\mathbb{N}^3$ such that $l=a$, $m=a+b$, $n=a+b+c$.
Thus, we have the following equality (all terms are positive, so order of summation can be chosen arbitrarily)
$$
\sum_{1<l\le m\le n}\frac{1}{5^l 3^m 2^n}=\sum_{a,b,c\in\mathbb{N}}\frac{1}{5^{a} 3^{a+b} 2^{a+b+c}}=\sum_{a,b,c\in\mathbb{N}}\frac{1}{30^a 6^b 2^c}=\sum_{a\in\mathbb{N}}\frac{1}{30^a}\cdot\sum_{b\in\mathbb{N}}\frac{1}{6^b}\cdot\sum_{c\in\mathbb{N}}\frac{1}{2^c}=
\\
=\frac{1}{30-1}\cdot\frac{1}{6-1}\cdot\frac{1}{2-1}\cdot=\frac{1}{145}.
$$
