Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$

Question

I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.

$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$

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$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{2^i3^j5^k}\\&^= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\frac{1}{2^i3^j}\frac{1/5^{j+1}}{4/5}\\&=\frac{1}{4}\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{1/15^{i+1}}{14/15}\\&=\frac{1}{56}\sum_{i=1}^{\infty}\frac{1}{30^{i}}=\frac{1}{56}\cdot\frac{1}{29}\end{aligned}$$

Answer 1

Wolfram Alpha yields $\dfrac{1}{1624}$, sounds like you are correct.

Answer 2

The index region of the sum \begin{align*} \sum_{\color{blue}{1\leq i <j<k}}\frac{1}{5^i3^j2^k}\tag{1} \end{align*} is specified by the inequality chain \begin{align*} 1\leq i <j<k \end{align*} which has $1$ as lower limit and $k-1$ as upper limit. We have two indices $i$ and $j$, which means we can write it as double sum as shown in the evaluation below.

We obtain \begin{align*} \color{blue}{\sum_{1\leq i<j<k}\frac{1}{5^{i}3^j2^k}} &=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\sum_{j=i+1}^{k-1}\frac{1}{3^j}\tag{2}\\ &=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\left(\frac{\left(\frac{1}{3}\right)^{i+1}-\left(\frac{1}{3}\right)^k}{1-\frac{1}{3}}\right)\tag{3}\\ &=\frac{1}{2^k}\sum_{i = 1}^{k-2}\frac{1}{5^i}\,\frac{1}{2}\left(\frac{1}{3^i}-\frac{1}{3^{k-1}}\right)\tag{4}\\ &= \frac{1}{2^{k+1}}\sum_{i=1}^{k-2}\frac{1}{15^i}- \frac{1}{2^{k+1}\,3^{k-1}}\sum_{i=1}^{k-2}\frac{1}{5^i}\\ &=\frac{1}{2^{k+1}}\left(\frac{\frac{1}{15}-\left(\frac{1}{15}\right)^{k-1}}{1-\frac{1}{15}}\right)-\frac{1}{2^{k+1}\,3^{k-1}}\left(\frac{\frac{1}{5}-\left(\frac{1}{5}\right)^{k-1}}{1-\frac{1}{5}}\right)\tag{5}\\ &=\frac{1}{2^{k+2}\cdot7}\left(1-\frac{1}{15^{k-2}}\right)-\frac{1}{2^{k+3 }\,3^{k-1}}\left(1-\frac{1}{5^{k-2}}\right)\\ &\,\,\color{blue}{=\frac{1}{2^{k+2}\cdot7}-\frac{1}{2^{k+3}\,3^{k-1}}+\frac{1}{2^{k+3}\,3^{k-1}\,5^{k-2}\cdot7}}\tag{6} \end{align*}

Comment:

• In (2) we factor out $\frac{1}{2^k}$ 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.

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Notes:

• Varying $k$ is not admissible since $k-1$ is an upper limit. Here $k$ is a free variable whereas the indices $i$ and $j$ are bound variables. This is different to the situation \begin{align*} \sum_{1\leq i <j<k\color{blue}{<\infty}}\frac{1}{5^{i}3^j2^k} \end{align*} where $k$ is an index bound by the upper limit $\infty$ and where $k$ varies between $j$ and $\infty$.

• The result (6) is also given by Wolfram Alpha with input

   sum(sum 1/(5^i*3^j*2^k), j=i+1..k-1),i=1..k-2
  

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.

Answer 3

Your answer is correct but there is a easier (to remember) way to derive the result.

To simplify the sum, the first thing one should do is get rid of the "$<$" constraint among the summation indices. If you introduce index $\ell, m$ such that $j = i + \ell$ and $k = j + m$, then the constraint $1 \le i < j < k$ simplifies to $1 \le i,\ell,m$. This will allow you to rewrite your iterated sum as a product of geometric series: $$\begin{align}\sum_{\substack{(i,j,k)\\1 \le i < j < k}}\frac{1}{2^i 3^j 5^k} &= \sum_{\substack{(i,j,k)\\1 \le i, \ell, m }}\frac{1}{2^i 3^{i+\ell} 5^{i+\ell + m}} =\sum_{i=1}^\infty \frac{1}{30^i}\sum_{\ell=1}^\infty\frac{1}{15^\ell}\sum_{m=1}^\infty\frac{1}{5^m}\\ &= \frac{\frac{1}{30}}{1-\frac1{30}}\frac{\frac{1}{15}}{1-\frac{1}{15}} \frac{\frac15}{1-\frac15} = \frac{1}{29}\frac{1}{14}\frac{1}{4} = \frac{1}{1624} \end{align} $$