# Evaluating $\sum_{(a,b,c)\in T}\frac{2^a}{3^b 5^c}$, for $T$ the set of all positive integer triples $(a,b,c)$ forming a triangle

Let $$T$$ be the set of all triples $$(a,b,c)$$ of positive integers for which there exist triangles with side lengths $$a$$, $$b$$, $$c$$. Express $$\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c}$$ as a rational number in lowest terms

I really don't know to start with this question. My solution: I have written equation of solution of triangle in form of $$a$$, $$b$$, and $$c$$; i.e, $$s/(s-a)$$, where $$s = (a+b+c)/2$$.

• apply Ravi substitution. – maveric Nov 8 '18 at 18:05
• maveric , i am really naive . Can you please elaborate or solve it if possible – Abhinov Singh Nov 8 '18 at 18:13
• put x= a+b-c y= b+c-a z= c+a-b – maveric Nov 8 '18 at 18:15

The main difficulties of this problem is how to handle the triangle inequalities: $$a+b > c, b+c > a, c+a > b$$ The beauty of Ravi substitution (mentioned by @maveric in comment) $$a = v + w, b = u + w, c = u + v$$ is under Ravi substitution,

$$a, b, c$$ satisfies the triangle inequalities if and only if $$u,v,w$$ are positive numbers.

If one want to run through all integer combinations of $$a,b,c$$ which give a triangle, one just run $$u,v,w$$ through all positive integer combinations or positive half-integer combinations $$(u,v,w) = \text{ one of }\begin{cases} (i+1, j+1, k+1 )\\ \\ (i+\frac12+i,j+\frac12,k+\frac12) \end{cases} \quad\text{ for }i,j, k\in \mathbb{N}$$

This transform the horrible sum to one over geometric series.

\begin{align} \sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c} = &\left(\frac{2^2}{15^2} + \frac{2}{15}\right)\sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty \frac{2^{j+k}}{3^{i+k}5^{i+j}}\\ = & \frac{34}{225}\sum_{i=0}^\infty \left(\frac{1}{15}\right)^i \sum_{j=0}^\infty \left(\frac{2}{5}\right)^j \sum_{k=0}^\infty \left(\frac{2}{3}\right)^k\\ = &\frac{34}{225}\left(\frac{1}{1 - \frac{1}{15}}\right) \left(\frac{1}{1 - \frac{2}{5}}\right)\left(\frac{1}{1 - \frac{2}{3}}\right)\\ = & \frac{17}{21} \end{align}

Naïve Solution

Note that $$(a,b,c)\in\mathbb{Z}_{>0}^3$$ is in $$T$$ iff $$|b-c|. Therefore, if $$S(p,q,r):=\sum_{(a,b,c)\in T}\,p^aq^br^c\,,$$ where $$p,q,r\in\mathbb{C}$$ with $$|qr|<1$$, $$|rp|<1$$, and $$|pq|<1$$, then $$S(p,q,r)=\sum_{b=1}^\infty\,q^b\,\sum_{c=1}^\infty\,r^c\,\sum_{a=|b-c|+1}^{b+c-1}\,p^a\,.$$ Without loss of generality, we may assume that $$p\neq \pm 1$$. That is, $$S(p,q,r)=\sum_{b=1}^\infty\,q^b\,\sum_{c=1}^\infty\,r^c\,\left(\frac{p^{|b-c|+1}-p^{b+c}}{1-p}\right)\,.$$ Consequentlty, $$S(p,q,r)=\small\frac{1}{1-p}\,\left(p\sum_{b=1}^\infty\,(qr)^b\,\sum_{c={b}}^\infty\,(pr)^{c-b}+p\,\sum_{b=1}^\infty\,(qr)^b\,\sum_{c=1}^{b-1}\,\left(\frac{p}{r}\right)^{b-c}-\sum_{b=1}^\infty\,(pq)^b\,\sum_{c=1}^\infty\,(pr)^c\right)\,.\tag{*}$$ If $$p\neq r$$, then $$S(p,q,r)=\small\frac{1}{1-p}\,\Biggl(p\,\left(\frac{qr}{1-qr}\right)\,\left(\frac{1}{1-pr}\right)+p\,\sum_{b=1}^\infty\,(qr)^b\,\frac{\frac{p}{r}-\left(\frac{p}{r}\right)^{b}}{1-\frac{p}{r}}-\left(\frac{pq}{1-pq}\right)\,\left(\frac{pr}{1-pr}\right)\Biggr)\,.$$ Ergo, for $$p\neq r$$, we have \begin{align}S(p,q,r)&=\frac{1}{1-p}\,\Biggl(p\,\left(\frac{qr}{1-qr}\right)\,\left(\frac{1}{1-pr}\right)+p\,\left(\frac{\frac{p}{r}}{1-\frac{p}{r}}\right)\,\left(\frac{qr}{1-qr}\right)\\&\phantom{aaaaa}-p\,\left(\frac{1}{1-\frac{p}{r}}\right)\,\left(\frac{pq}{1-pq}\right)-\left(\frac{pq}{1-pq}\right)\,\left(\frac{pr}{1-pr}\right)\Biggr)\,.\end{align} Simplifying the expression, we get $$S(p,q,r)=\frac{pqr\,(1+pqr)}{(1-qr)\,(1-rp)\,(1-pq)}\tag{#}$$ when $$p\neq r$$..

If $$p=r$$, then you can use continuity to conclude that (#) holds. Alternatively, from (*), we have $$S(p,q,p)=\small\frac{1}{1-p}\,\left(p\sum_{b=1}^\infty\,(pq)^b\,\sum_{c={b}}^\infty\,(p^2)^{c-b}+p\,\sum_{b=1}^\infty\,(b-1)\,(pq)^b-\sum_{b=1}^\infty\,(pq)^b\,\sum_{c=1}^\infty\,(p^2)^c\right)\,.$$ That is, $$S(p,q,p)=\small\frac{1}{1-p}\,\Biggl(p\,\left(\frac{pq}{1-pq}\right)\,\left(\frac{1}{1-p^2}\right)+p\,\left(\frac{pq}{1-pq}\right)^2-\left(\frac{pq}{1-pq}\right)\,\left(\frac{p^2}{1-p^2}\right)\Biggr)\,.$$ Upon simplifying, we get $$S(p,q,p)=\frac{p^2q\,(1+p^2q)}{(1-p^2)\,(1-pq)^2}\,,$$ which agrees with (#).

Now, in this particular problem, $$p=2$$, $$q=\dfrac{1}{3}$$, and $$r=\dfrac{1}{5}$$. Therefore, by (#), $$S\left(2,\frac13,\frac15\right)=\frac{17}{21}\,.$$ I was tempted to use Ravi's substitutions, but wanted to illustrate that a direct approach is not that bad.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$

$$\ds{\sum_{\pars{a,b,c}\ \in\ T} {2^{a} \over 3^{b}5^{c}}:\ {\LARGE ?}\qquad \\[2mm] T \equiv \braces{\pars{a,b,c}\ \mid\ a,b,c \in \mathbb{N}_{\geq 1}\ \mbox{and}\ \verts{b - c} < a < b + c}}$$

I'll evaluate \begin{align} &\bbox[10px,#ffd]{\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}} \bracks{\vphantom{\large A}\verts{b - c} < a < b + c}} \\[5mm] = &\ \sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}} \bracks{b < c}\bracks{\vphantom{\large A}c - b < a < c + b} \\[2mm] + &\ \sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty}{x^{a} \over \pars{yz}^{b}} \bracks{b = c}\bracks{\vphantom{\large A}a < 2b} \\[2mm] + &\ \sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}} \bracks{b > c}\bracks{\vphantom{\large A}b - c < a < b + c} \\[5mm] = &\ \overbrace{\sum_{b = 1}^{\infty}{1 \over y^{b}} \sum_{c = b + 1}^{\infty}{1 \over z^{c}} \sum_{a = c - b + 1}^{c + b - 1}x^{a}} ^{\mbox{See expression}\ {\large\eqref{2}}}\ +\ \overbrace{\sum_{b = 1}^{\infty}{1 \over \pars{yz}^{b}} \sum_{a = 1}^{2b - 1}x^{a}} ^{\mbox{See expression}\ {\large\eqref{3}}} \\[2mm] + &\ \underbrace{\quad\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}} \bracks{b > c}\bracks{\vphantom{\large A}b - c < a < b + c}\quad} _{\ds{\mbox{It's equal to the}\ first\ term\ \mbox{under the exchange}\ y \leftrightarrow z}} \label{1}\tag{1} \end{align}

The RHS first term becomes: \begin{align} &\bbox[#ffd,10px]{\sum_{b = 1}^{\infty}{1 \over y^{b}} \sum_{c = b + 1}^{\infty}{1 \over z^{c}} \sum_{a = c - b + 1}^{c + b - 1}x^{a}} = \sum_{b = 1}^{\infty}{1 \over y^{b}} \sum_{c = b + 1}^{\infty}{1 \over z^{c}}\, x^{c - b + 1}\,{x^{2b - 1} - 1 \over x - 1} \\[5mm] = &\ \sum_{b = 1}^{\infty}{1 \over y^{b}} \,{x^{b} - x^{-b + 1} \over x - 1} \sum_{c = b + 1}^{\infty}\pars{x \over z}^{c} \\[5mm] = &\ {1 \over x - 1}\sum_{b = 1}^{\infty} \,\bracks{\pars{x \over y}^{b} - {x \over \pars{xy}^{b}}} {\pars{x/z}^{b + 1} \over 1 - x/z} \\[5mm] = &\ {z \over \pars{x - 1}\pars{z - x}}\sum_{b = 1}^{\infty} \,\bracks{{x \over z}\pars{x^{2} \over yz}^{b} - {x^{2} \over z} \pars{1 \over yz}^{b}} \\[5mm] = &\ {z \over \pars{x - 1}\pars{z - x}} \,\bracks{{x \over z}{x^{2}/\pars{yz} \over 1 - x^{2}/\pars{yz}} - {x^{2} \over z}{1/\pars{yz} \over 1 - 1/\pars{yz}}} \\[5mm] = &\ \bbx{{x^{3} \over \pars{x - 1}\pars{z - x}\pars{yz - x^{2}}} - {x^{2} \over \pars{x - 1}\pars{z - x}\pars{yz - 1}}} \label{2}\tag{2} \end{align}
The RHS second term in \eqref{1} is given by \begin{align} &\bbox[10px,#ffd]{\sum_{b = 1}^{\infty}{1 \over \pars{yz}^{b}}\, \sum_{a = 1}^{2b - 1}x^{a}} = \sum_{b = 1}^{\infty}\pars{1 \over yz}^{b}\, x\,{x^{2b - 1} - 1 \over x - 1} \\[5mm] = &\ {1 \over x - 1}\bracks{% \sum_{b = 1}^{\infty}\pars{x^{2} \over yz}^{b} - x\sum_{b = 1}^{\infty}\pars{1 \over yz}^{b}} \\[5mm] = &\ {1 \over x - 1}\bracks{% {x^{2}/\pars{yz} \over 1 - x^{2}/\pars{yz}} - x\,{1/\pars{yz} \over 1 - 1/\pars{yz}}} \\[5mm] = &\ \bbx{{1 \over x - 1}\pars{% {x^{2} \over yz - x^{2}} -{x \over yz - 1}}} \label{3}\tag{3} \end{align}
The last term in \eqref{1} RHS is found, as pointed out above, with the exchange $$\ds{y \leftrightarrow z}$$ in \eqref{2}.

The final result is given by

$$\bbx{\bbox[10px,#ffd]{\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}} \bracks{\vphantom{\large A}\verts{b - c} < a < b + c}} = {x\pars{x + yz} \over \pars{x - y}\pars{x - z}\pars{yz - 1}}}$$

When $$\ds{x = 2\,,\ y = 3}$$ and $$\ds{z = 5}$$, the above expression is reduced to:

$$\bbx{\sum_{\pars{a,b,c}\ \in\ T} {2^{a} \over 3^{b}5^{c}} = {17 \over 21}}$$

Not an answer but too much for a comment.

Case 1. $$a=b=c, S_1=\sum_{n=1}^\infty{\left(\cfrac 2{15}\right)^n}$$ That's the easy case;

Case 2. $$\displaystyle a=b\ne c, S_2=\sum_{(a,c)\in U}\left(\cfrac{2^a}{3^a5^c}+\cfrac{2^a}{3^c5^a}+\cfrac{2^c}{3^a5^a}\right)=\sum_{i=1}^\infty\left(\cfrac{2^i}{3^i}\left(\sum_{j=1}^{2i-1, j\ne i}\cfrac 1{5^j}\right)+\cdots+\cdots\right)$$,

with $$U$$ the set of integer tuples that can form an isosceles triangle with the first entity the sides.

Case 3. $$a< b < c$$, $$S_3=$$sum of 6 terms (Permutation of 3) for each triple of $$a,b,c$$ satisfying $$a+b>c$$

$$S_3$$ needs further investigation. $$S_1, S_2$$ is almost there. Total=$$S_1+S_2+S_3$$