Section 6

Distribution of the Signs of the Terms

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Problem 1.138

Assume that pn>0p_n>0, p1≥p2≥p3≥⋯p_1 \geq p_2 \geq p_3 \geq \cdots and that the series

p1+p2+p3+⋯+pn+⋯ p_1+p_2+p_3+\cdots +p_n+\cdots

is divergent and the series

ε1p1+ε2p2+ε3p3+⋯+εnpn+⋯ , \varepsilon _1 p_1 + \varepsilon _2 p_2 + \varepsilon _3 p_3 + \cdots + \varepsilon _n p_n + \cdots ,

where εk\varepsilon_k is −1-1 or +1+1, is convergent. If under these conditions a certain percentage of terms is positive then it is 50%50\%. More precisely:

lim inf⁡n→∞ε1+ε2+⋯+εnn≤0≤lim sup⁡n→∞ε1+ε2+⋯+εnn. \liminf _{n \rightarrow \infty } \frac{\varepsilon _1+\varepsilon _2+\cdots +\varepsilon _n}{n} \leq 0 \leq \limsup _{n \rightarrow \infty } \frac{\varepsilon _1+\varepsilon _2+\cdots +\varepsilon _n}{n} .
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Remark.
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E. Cesàro: Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur Ser. 4, Vol. 4, 2nd Sem. p. 133 (1888); J. Bagnera: Darboux Bull. Ser. 2, Vol. 12, p. 227 (1888). Cf.also G. H. Hardy: Mess. Math. Ser. 2, Vol. 41, p. 17 (1911); H. Rademacher: Math. Z. Vol. 11, pp. 276-288 (1921).

Problem 1.139

Suppose pn>0p_n>0, p1≥p2≥p3≥⋯p_1 \geq p_2 \geq p_3 \geq \cdots and that the series

ε1p1+ε2p2+ε3p3+⋯+εnpn+⋯ , \varepsilon _1 p_1 + \varepsilon _2 p_2 + \varepsilon _3 p_3 + \cdots + \varepsilon _n p_n + \cdots ,

where εk\varepsilon_k is −1-1 or +1+1, is convergent. Then

lim⁡n→∞(ε1+ε2+ε3+⋯+εn) pn=0. \lim _{n \rightarrow \infty } (\varepsilon _1+\varepsilon _2+\varepsilon _3+\cdots +\varepsilon _n) \, p_n = 0 .
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Remark.
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Notice the two well-known extreme cases ε1=ε2=ε3=⋯\varepsilon_1=\varepsilon_2=\varepsilon_3=\cdots and ε1=−ε2=ε3=−ε4=⋯\varepsilon_1=-\varepsilon_2=\varepsilon_3=-\varepsilon_4=\cdots. E. Lasker.