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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 :/

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    $\begingroup$ You seem not to respect the constraints $\ell<m$, $m<n$. $\endgroup$
    – Nicolas
    Dec 20, 2020 at 10:07
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    $\begingroup$ 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 $\endgroup$
    – user822140
    Dec 20, 2020 at 10:19
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    $\begingroup$ I would not execute the sum. But I may evaluate it. $\endgroup$
    – GEdgar
    Dec 20, 2020 at 12:58
  • $\begingroup$ 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 $\sum_{0\leq l<n}q^l=\sum_{l=0}^{n-1}q^l=\frac{1-q^n}{1-q}$. $\endgroup$ Dec 29, 2020 at 15:59

3 Answers 3

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  1. 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 $$\begin{align}\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} &= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \left(\frac{1}{2^m}\right) \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{6^m} \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell}\left(\frac{1}{5\cdot 6^\ell}\right) \\& = \frac{1}{5}\sum_{\ell=1}^\infty \frac{1}{30^\ell} \\&= \frac{1}{5}\cdot\frac{1}{29} \\&= \frac{1}{145}\end{align} $$


  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.

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  • $\begingroup$ @VIVD: Note the difference between the index regions: $1\leq \ell<m<n$ and $1\leq \ell<m<n<\infty$. $\endgroup$ Dec 21, 2020 at 13:15
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    $\begingroup$ @MarkusScheuer Yes, you are right. Since the question was quite ambiguous, I added a comment about the second case, as well. $\endgroup$
    – VIVID
    Dec 21, 2020 at 13:23
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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) $$ \sum_{1<l\le m\le n}\frac{1}{5^l 3^m 2^n}=\sum_{a,b,c\in\mathbb{N}}\frac{1}{5^{a} 3^{a+b} 2^{a+b+c}}=\sum_{a,b,c\in\mathbb{N}}\frac{1}{30^a 6^b 2^c}=\sum_{a\in\mathbb{N}}\frac{1}{30^a}\cdot\sum_{b\in\mathbb{N}}\frac{1}{6^b}\cdot\sum_{c\in\mathbb{N}}\frac{1}{2^c}= \\ =\frac{1}{30-1}\cdot\frac{1}{6-1}\cdot\frac{1}{2-1}\cdot=\frac{1}{145}. $$

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

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