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Statement

Example 1 (p. 132). Let aa and bkb_k be integers with a>1a>1 and ∑∣bk∣a−2k<∞\sum|b_k|a^{-2^k}<\infty, and put nk=a2k+bkn_k=a^{2^k}+b_k. Then ∑1/nk\sum1/n_k is irrational.

The paper marks the example with its reference [1], printed as J. W. Golomb, A Special Case, On the sum of the reciprocals of the Fermat numbers and related irrationalities, Canad. J. Math. 15 (1963), 475--478; the article is by S. W. Golomb. The case a=2a=2, bk=1b_k=1 is the sum of the reciprocals of the Fermat numbers 22k+12^{2^k}+1.

Proof pointer

Pp. 132--133. The convergence hypothesis gives nk2/nk+1→1n_k^2/n_{k+1}\to1, which is condition (i) of Theorem 1, and Nk∗/nk+1=a−1∏l=1k(1+bla−2l)/(1+bk+1a−2k+1)N_k^*/n_{k+1}=a^{-1}\prod_{l=1}^k(1+b_la^{-2^l})/(1+b_{k+1}a^{-2^{k+1}}) is bounded, which gives condition (ii) since Nk≤Nk∗N_k\le N_k^*. A rational sum would then force the recurrence, which reads bk+1=2a2kbk+bk2−a2k−bk+1b_{k+1}=2a^{2^k}b_k+b_k^2-a^{2^k}-b_k+1. Then bk≠0b_k\ne0 forces ∣bk+1∣>a2k|b_{k+1}|>a^{2^k} for large kk (11), and bk=0b_k=0 forces bk+1=1−a2k≠0b_{k+1}=1-a^{2^k}\ne0; iterating (11) keeps ∣bk+l∣a−2k+l|b_{k+l}|a^{-2^{k+l}} above a−2ka^{-2^k}, so it does not tend to 00, against the hypothesis.

Dependencies

Theorem 1.

Source. P. Erdős and E. G. Straus, On the irrationality of certain Ahmes series, J. Indian Math. Soc. (N.S.) 27 (1964), 129--133; the edition read is named on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image of p. 132 and the proof on pp. 132--133 for its structure. Nothing here is independently reviewed.

Bears on

  • Problem 243: an instance. The sequences of Example 1 satisfy the problem's hypothesis an/an−12→1a_n/a_{n-1}^2\to1, and the proof shows they do not satisfy the recurrence for all large nn; their reciprocal sums are irrational, so the problem's statement holds for this family.