# How to read and execute $\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$

How to read and execute this sum? $$\sum_{1 \leq \ell

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 But this does not agree with the answer :/

• You seem not to respect the constraints $\ell<m$, $m<n$. Dec 20, 2020 at 10:07
• Your error is a big one. It is like saying $a_1b_1c_1 + a_2b_2c_2=(a_1+a_2)(b_1+b_2)(c_1+c_2)$.An anology- There are asking for 3 nested loops whereas you are taking 3 seperate loops and multiplying them
– user822140
Dec 20, 2020 at 10:19
• I would not execute the sum. But I may evaluate it. Dec 20, 2020 at 12:58
• Note $n$ is not an index, but $n−1$ is an upper limit instead. An assumption that $n$ is not fixed is not admissible. Consider for instance $\sum_{0\leq l<n}q^l=\sum_{l=0}^{n-1}q^l=\frac{1-q^n}{1-q}$. Dec 29, 2020 at 15:59

## 3 Answers

1. Assuming $$n$$ is not fixed:

Rewrite as triple sum $$\sum_{1 \leq \ell 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}

1. 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.

• @VIVD: Note the difference between the index regions: $1\leq \ell<m<n$ and $1\leq \ell<m<n<\infty$. Dec 21, 2020 at 13:15
• @MarkusScheuer Yes, you are right. Since the question was quite ambiguous, I added a comment about the second case, as well. Dec 21, 2020 at 13:23

We will assume that $$n$$ is not fixed. Firstly, note that for any triple $$(l,m,n)\in\mathbb{N}^3$$ such that $$l there is exactly one triple $$(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

• The assumption that $n$ is not fixed is not admissible in the current situation. Regards, Dec 22, 2020 at 13:42

The index region of the sum \begin{align*} \sum_{\color{blue}{1\leq \ell is specified by the inequality chain \begin{align*} 1\leq \ell 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

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 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.