= Solution
Write $A_N(f)=(N+1)^{-1}\sum_{n=0}^N f(e^{in\xi})$. For a monomial $z^k$, $k\in\mathbb Z$, the <geometric series> gives
$$
A_N(z^k)=\begin{cases}1,&e^{ik\xi}=1,\\ \displaystyle\frac{1-e^{ik(N+1)\xi}}{(N+1)(1-e^{ik\xi})},&e^{ik\xi}\ne1.\end{cases}
$$
Thus every such average converges, and by <linearity> so does the average of every <trigonometric polynomial>. Also $|A_N(f)|\leq\|f\|_\infty$ uniformly in $N$.
Given $\varepsilon>0$, choose a <trigonometric polynomial> $P$ with $\|f-P\|_\infty<\varepsilon$. Then
$$
|A_N(f)-A_M(f)|\leq2\varepsilon+|A_N(P)-A_M(P)|.
$$
The last term tends to zero as $N,M\to\infty$. The scalar sequence is Cauchy and therefore \b[the limit exists for every continuous $f$ and every real $\xi$].
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