Sum of reciprocals of odd numbers that add up to $2$

Question

It is well known that certain sums of reciprocals add up to $2$:

• the reciprocals of powers of $2$:

$$ {1 \over 1}+{1 \over 2}+{1 \over 4}+{1 \over 8}+{1 \over 16}+{1 \over 32}+\cdots =2, $$

• the reciprocals of the triangular numbers:

$$ {1 \over 1}+{1 \over 3}+{1 \over 6}+{1 \over 10}+{1 \over 15}+{1 \over 21}+\cdots =2, $$

• $\ldots$

Are there any sums such that $\sum_{i=0}^{\infty} s_i^{-1}=2$ for which all the $s_i$ are all distinct odd positive integers?

Similarly, are there any sums such that $\exists k\in\mathbb{Z}, k>0$ such that $\sum_{i=0}^{k} s_i^{-1}=2$ for which all the $s_i$ are all distinct odd positive integers?

Answer 1

If you allow $\frac{1}{1}$, which you seem to, judging by your question:

Try adding the reciprocals of $1, 3, 5, 7, 9, 15, 21, 27, 35, 63, 105, 135$.

Found by choosing an odd abundant number, namely $945$ and finding a sum of factors that adds to $2\cdot 945$:

$945+315+189+135+105+63+45+35+27+15+9+7=2\cdot 945$.

Then divide this equation by $945$, reducing the individual fractions on the left.

Answer 2

We can use a variation of the greedy algorithm to find such a sequence in a deterministic way.

First, we add up the reciprocal odd numbers until their sum exheeds $2$:

$$1+\frac{1}{3}+\dots+\frac{1}{13}<2$$

$$1+\frac{1}{3}+\dots+\frac{1}{13}+\frac{1}{15}>2$$

Thus, we start with:

$$S_{7}=\sum_{n=1}^{7} \frac{1}{2n-1}$$

1) After subtracting the sum we are left with:

$$2-S_{7}=\frac{2021}{45045}$$

To find the next term we need to find an integer $m_1$ such that:

$$\frac{1}{2m_1-1} < \frac{2021}{45045} < \frac{1}{2m_1-3}$$

Turns out $m_1=12$, because:

$$\frac{1}{23} < \frac{2021}{45045} < \frac{1}{21}$$

2) After subtracting, we are left with:

$$2-S_{7}-\frac{1}{23}=\frac{1438}{1036035}$$

Because:

$$\frac{1}{721} < \frac{1438}{1036035} < \frac{1}{719}$$

We have $m_2=361$.

3) After subtracting, we are left with:

$$2-S_{7}-\frac{1}{23}-\frac{1}{721}=\frac{109}{106711605}$$

We find that:

$$\frac{1}{979007} < \frac{109}{106711605} < \frac{1}{979005}$$

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So for now we have:

$$2=\sum_{n=1}^{7} \frac{1}{2n-1}+\frac{1}{23}+\frac{1}{721}+\frac{1}{979007}+\dots$$

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Some comments. We can continue in the same fashion getting larger and larger denominators. I'm not sure if the sequence will be finite or not.

If we were allowed to use all integers, then the sequence will be finite, because $2$ is a rational number. Also we would know that denominators would grow approximately as $a_{n+1} > a_n^2-a_n$, which simplifies the search for the next one.

In the case of odd integers things become a little more complicated. Whether all rational numbers with odd denominators produce a finite odd greedy expansion is unknown to this day.

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If we were allowed to change signs, as luck would have it, we can find a relatively short sequence:

$$2=\sum_{n=1}^{7} \frac{1}{2n-1}+\frac{1}{23}+\frac{1}{715}-\frac{1}{94185}$$