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Charles
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Is there a standard definition for a lacunary sequence?

Suppose $0 < a_1 < a_2 < \cdots.$

I've read two papers using the term recently. One requires $$ \lim_n\frac{a_{n+1}}{a_n}>1 $$$$ \liminf_n\frac{a_{n+1}}{a_n}>1 $$ while the other only requires $$ \lim_na_{n+1}-a_n=+\infty. $$

The two differ, of course: $a_1=1,\ a_{n+1}=a_n+\sqrt{a_n}$ has a ratio that tends to 1 but a difference that diverges.

Further, the EOM entry for lacunary sequence is different from both (a finite form of the first): $$ \frac{a_{n+1}}{a_n}\ge\lambda>1. $$

Is there a standard definition for a lacunary sequence?

Suppose $0 < a_1 < a_2 < \cdots.$

I've read two papers using the term recently. One requires $$ \lim_n\frac{a_{n+1}}{a_n}>1 $$ while the other only requires $$ \lim_na_{n+1}-a_n=+\infty. $$

The two differ, of course: $a_1=1,\ a_{n+1}=a_n+\sqrt{a_n}$ has a ratio that tends to 1 but a difference that diverges.

Further, the EOM entry for lacunary sequence is different from both (a finite form of the first): $$ \frac{a_{n+1}}{a_n}\ge\lambda>1. $$

Is there a standard definition for a lacunary sequence?

Suppose $0 < a_1 < a_2 < \cdots.$

I've read two papers using the term recently. One requires $$ \liminf_n\frac{a_{n+1}}{a_n}>1 $$ while the other only requires $$ \lim_na_{n+1}-a_n=+\infty. $$

The two differ, of course: $a_1=1,\ a_{n+1}=a_n+\sqrt{a_n}$ has a ratio that tends to 1 but a difference that diverges.

Further, the EOM entry for lacunary sequence is different from both (a finite form of the first): $$ \frac{a_{n+1}}{a_n}\ge\lambda>1. $$

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Charles
  • 9.1k
  • 1
  • 38
  • 76

Lacunary sequence

Is there a standard definition for a lacunary sequence?

Suppose $0 < a_1 < a_2 < \cdots.$

I've read two papers using the term recently. One requires $$ \lim_n\frac{a_{n+1}}{a_n}>1 $$ while the other only requires $$ \lim_na_{n+1}-a_n=+\infty. $$

The two differ, of course: $a_1=1,\ a_{n+1}=a_n+\sqrt{a_n}$ has a ratio that tends to 1 but a difference that diverges.

Further, the EOM entry for lacunary sequence is different from both (a finite form of the first): $$ \frac{a_{n+1}}{a_n}\ge\lambda>1. $$