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

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.

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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. – Sangchul Lee Dec 20 '12 at 23:58
Yeah I think the subscript 4 is supposed to be 2 in the book. – Christmas Bunny Dec 21 '12 at 17:52

Let $(x_n)_{n\geqslant1}$ denote a bounded sequence, $s_n=\sum\limits_{k=1}^nx_k$ and $u_n=\tfrac1ns_n$. Then $(u_n)_{n\geqslant1}$ converges (to a limit $\ell$) if and only if $(u_{n^2})_{n\geqslant1}$ converges (to the same limit $\ell$).
Only one direction needs proof, hence one assumes that $(u_{n^2})_{n\geqslant1}$ converges. One can assume without loss of generality that $|u_n|\leqslant C$ for every $n$ and that $\ell=0$. Then, for every $n\geqslant1$, there exists $k\geqslant1$ such that $k^2\leqslant n\lt (k+1)^2$, and $$|u_n|\leqslant\tfrac1n|s_{k^2}|+2kC\tfrac1n\leqslant|u_{k^2}|+\tfrac2kC.$$ When $n\to\infty$, $k\to\infty$ (since $k\gt\sqrt{n}-1$) hence $|u_{k^2}|\to0$ and $\tfrac2k\to0$, which proves that $|u_n|\to0$.
The sum in the second limit appears to only have $n$ terms, not $n^2$ - it iterates over only the square elements of $a_n$. – Steven Stadnicki Dec 21 '12 at 0:07
@did: why can we assume WLOG $\alpha=0$? – Christmas Bunny Dec 21 '12 at 17:53