# Summation of infinite Series where triple ranges over set of possible triangle lengths.

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 $$\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c}$$ as a rational number in lowest terms.

I don't really know how to start!.

The answer is $17/21$. For fixed $b,c$, there is a triangle of side lengths $a,b,c$ if and only if $|b-c|<a<b+c$. It follows that the desired sum is $$S = \sum_{b,c} \frac{1}{3^b5^c} \left(\sum_{a=|b-c|+1}^{b+c-1} 2^a \right) = \sum_{b,c} \frac{2^{b+c}-2^{|b-c|+1}}{3^b5^c}.$$ We write this as $S = S_1+S_2$ where $S_1$ sums over positive integers $b,c$ with $b\leq c$ and $S_2$ sums over $b>c$. Then \begin{align*} S_1 &= \sum_{b=1}^\infty \sum_{c=b}^\infty \frac{2^{b+c}-2^{c-b+1}}{3^b 5^c} \\ &= \sum_{b=1}^\infty \left( \left( \left(\frac{2}{3}\right)^b-\frac{2}{6^b} \right) \sum_{c=b}^\infty \left(\frac{2}{5} \right)^c \right) \\ &= \sum_{b=1}^\infty \left( \left(\frac{2}{3}\right)^b-\frac{2}{6^b} \right) \frac{5}{3} \left( \frac{2}{5} \right)^b \\ &= \sum_{b=1}^\infty \left( \frac{5}{3} \left(\frac{4}{15}\right)^b - \frac{10}{3} \left(\frac{1}{15}\right)^b \right) \\ &= \frac{85}{231}. \end{align*} Similarly, \begin{align*} S_2 &= \sum_{c=1}^\infty \sum_{b=c+1}^\infty \frac{2^{b+c}-2^{b-c+1}}{3^b 5^c} \\ &= \sum_{c=1}^\infty \left( \left( \left(\frac{2}{5}\right)^c-\frac{2}{10^c} \right) \sum_{b=c+1}^\infty \left(\frac{2}{3} \right)^b \right) \\ &= \sum_{c=1}^\infty \left( \left(\frac{2}{5}\right)^c-\frac{2}{10^c} \right) 3 \left( \frac{2}{3} \right)^{c+1} \\ &= \sum_{c=1}^\infty \left( 2 \left(\frac{4}{15}\right)^c - 4 \left(\frac{1}{15}\right)^c \right) \\ &= \frac{34}{77}. \end{align*} We conclude that $S = S_1+S_2 = \frac{17}{21}$.
• $|b-c| <a<|b+c|$ does the trick. Wow.Thanks – user567182 Jun 15 '18 at 2:30