Prove that for any integer $n > 3$ there exist positive integers $a_1 > \dots > a_n$ such that $1 = \frac{1}{a_1} + \dots + \frac{1}{a_n}$

Question

I have tried to approach this problem in a variety of ways, like triyng to construct a solution, looking at telescoping series, geometrically... But I have been unable to solve it. I would appreciate any hint. Thanks.

Answer 1

Since OP has solved the question, here is another solution to the question (besides using Sylvester's sequence). Observe for $n\ge3$:

$$1 - \frac1{2^{n-2}}=\frac12+\frac14+\dots+\frac1{2^{n-2}}$$

and

$$\frac1{2^{n-2}} = \frac 1{3 \times 2^{n-1}}+\frac1{3\times 2^{n-2}}$$

and thus

$$1 = \frac12+\frac14+\dots+\frac1{2^{n-2}}+\frac 1{3 \times 2^{n-1}}+\frac1{3\times 2^{n-2}}$$

which consists of exactly $n$ terms.