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This is an assert in the book "Problems from the book" by Titu.

Suppose the Sylvester sequence is defined by $u_1 = 2, u_{n+1} = u_n^2 - u_n + 1$, so $(u_1, u_2,...) = (2,3,7,43,...)$

Suppose there are $N$ positive integers $a_1,a_2,\dots,a_N$ satisfing $\sum_{i=1}^N \frac{1}{a_i} < 1$, prove that $\sum_{i=1}^N \frac{1}{a_i} \leq \sum_{i=1}^N \frac{1}{u_i}$

I don't find a clue about this, even I find the name of "Sylvester sequence".

EggTart
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    It is in "Problems from the book" Ch. 8, Page 161. "Proofs from the book" is another book. – River Li Nov 14 '23 at 23:57
  • Thanks, just corrected it. – EggTart Nov 15 '23 at 07:11
  • Can you find the original reference that Erdos asserted this result? – River Li Nov 16 '23 at 01:24
  • https://en.wikipedia.org/wiki/Sylvester%27s_sequence gives the following references: Curtiss, D. R. (1922). "On Kellogg's diophantine problem". American Mathematical Monthly. 29 (10): 380–387. doi:10.2307/2299023. JSTOR 2299023; Miller, G. A. (1919). "Groups possessing a small number of sets of conjugate operators". Transactions of the American Mathematical Society. 20 (3): 260–270. doi:10.2307/1988867. JSTOR 1988867. Rosenman, Martin; Underwood, F. (1933). "Problem 3536". American Mathematical Monthly. 40 (3): 180–181. doi:10.2307/2301036. JSTOR 2301036. (continued) – Gerry Myerson Nov 29 '23 at 06:46
  • (continued) Salzer, H. E. (1947). "The approximation of numbers as sums of reciprocals". American Mathematical Monthly. 54 (3): 135–142. doi:10.2307/2305906. JSTOR 2305906. MR 0020339. and Soundararajan, K. (2005). "Approximating 1 from below using n Egyptian fractions". arXiv:math.CA/0502247 (that last one is also in the answer by user EggTart). – Gerry Myerson Nov 29 '23 at 06:47
  • Related question: https://math.stackexchange.com/questions/2905659/show-sylvester-sequence-is-the-smallest-solution-with-n-terms-to-sum-of-unit-fra – Gerry Myerson Nov 29 '23 at 06:55
  • Any thoughts about the answer that has been posted, or about the links I gave, Egg? – Gerry Myerson Dec 01 '23 at 01:41
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    I’m voting to close this question because OP has disengaged. – Gerry Myerson Dec 02 '23 at 05:26

1 Answers1

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I just find this short note which is really helpful for this problem. The crucial trick here is to wisely use Muirhead inequality combining an induction.

For the proof, see https://arxiv.org/pdf/math/0502247.pdf

The most important lemma used in the answer is the following result: if some positive number satisfies $x_1 \geq x_2 \geq \dots \geq x_n, y_1 \geq y_2 \geq \dots\geq y_n$, $\Pi_{i=1}^k x_i \geq \Pi_{i=1}^k y_i$ for every $k \geq 1$, then $\sum_{i=1}^n x_i \geq \sum_{i=1}^n y_i$.

And the virtue of sylvester sequence is that $\sum_{i=1}^n \frac{1}{a_i} = 1 - \frac{1}{\Pi_{i=1}^n a_i}$, which means that if we want $\sum_{i=1}^n\frac{1}{b_i}$ to be sufficient close to 1, then you must have a large enough $\Pi_{i=1}^n b_i$.

EggTart
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Nov 15 '23 at 08:12