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Show that for any sequence $a_1,a_2,...$ of real numbers, the two conditions

$\lim_{n\to\infty}\frac{\exp(ia_1)+\exp(ia_2)+...+\exp(ia_n)}{n}=\alpha$

and

$\lim_{n\to\infty}\frac{\exp(ia_1)+\exp(ia_4)+...+\exp(ia_{n^2})}{n^2}=\alpha$

are equivalent.

  • 1
    I can hardly believe that it is true, since if we put $a_n = 0$ identically, the former limit is 1 while the latter is 0.2012-12-20
  • 0
    Yeah I think the subscript 4 is supposed to be 2 in the book.2012-12-21

1 Answers 1