# Sum of reciprocals of powers of 3 equals 3/8?

Let $$A=\{ \sum_{i=1}^\infty \frac{x_i}{3^i}: x_i = 0 \,\text{ or }\, 2 \}$$, show that $$\frac{3}{4} \in A$$

My attempt:

Let $$2\bigl(\frac{1}{3^{n_1}}+\frac{1}{3^{n_2}}+\frac{1}{3^{n_3}} + \dotsm\bigr) = \frac{3}{4}$$

$$\Rightarrow 2(\frac{3p+1}{3^{n_1}}) = \frac{3}{4}$$, if $$n_1>n_2>n_3>...$$

$$\Rightarrow 8(3p+1) = 3^{n_1+1}$$

$$\Rightarrow 8 = 3q$$, where q is an integer.

Which is impossible.

• 3=3/8?? typo i guess! – Albus Dumbledore Sep 6 '20 at 10:54
• @Quantum No. The title is (sum of reciprocals of poweres of $3$) = $3/8$. – José Carlos Santos Sep 6 '20 at 10:55
• One problem I see with your proof is that you denote $n_1$ to be the biggest power. But you can't do that, since they could be infinitely many and not have a biggest. – Todor Markov Sep 6 '20 at 10:57
• math.stackexchange.com/questions/289803/… See John's Answer. – Sumanta Sep 6 '20 at 10:59
• If you don't see yet why your proof is incorrect, you can try to exaggerate it a bit: $\pi=3+\frac{1}{10}+\frac{4}{10^2}+\frac{1}{10^3}+\frac{5}{10^4}+\ldots$. If what you've attempted was possible, you would get $\pi=\frac{q}{10^{n}}$ for some $n$, i.e. $\pi$ would be rational. – Stinking Bishop Sep 6 '20 at 11:08

The greedy algorithm works here. Since $$2/3 < 3/4$$, we pick $$x_1 = 2$$. Then the remainder is $$1/12$$, which we compare against $$2/9$$ and is smaller, so $$x_2 = 0$$. The next power of $$3$$ is $$27$$, and we have $$2/27 < 2/24 = 1/12$$, so $$x_3 = 2$$, leaving a remainder of $$1/108$$. We next observe that $$\frac{1}{108} = \frac{3}{4} \cdot \frac{1}{3^4},$$ so it turns out that \begin{align} \frac{3}{4} &= \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} \cdot \frac{3}{4} \\ &= \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} \left( \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} \cdot \frac{3}{4} \right) \\ &= \frac{2}{3^1} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{0}{3^4} + \frac{2}{3^5} + \frac{0}{3^6} + \frac{2}{3^7} + \frac{1}{3^8} \cdot \frac{3}{4} \\ &= \cdots \\ \end{align} and it becomes apparent that the pattern is $$x_{2k+1} = 2, \quad x_{2k+2} = 0.$$ We can formally verify this by noting that our sum is now $$\sum_{k=0}^\infty \frac{2}{3^{2k+1}} + \frac{0}{3^{2k+2}} = \frac{2}{3} \sum_{k=0}^\infty \frac{1}{9^k} = \frac{2/3}{1 - 1/9} = \frac{3}{4}.$$ The astute observer will note that the sequence of $$x_i$$ corresponds to the base-$$3$$ or ternary digit expansion of the rational $$3/4$$; i.e., $$\frac{3}{4} = (0.202020\ldots)_3.$$ We went through two cycles of the expansion instead of just one: we could have noted that $$\frac{3}{4} = \frac{2}{3^1} + \frac{0}{3^2} + \frac{1}{3^2} \cdot \frac{3}{4}.$$
To answer the question in the body: $$9 \cdot \frac 34 = 6 + \frac 34 \implies \frac 34 = 0.202020\cdots \text{ in base } 3$$