Define $b_n = \int_0^{\pi} \cos (nx).\sqrt{x} ~ dx $.
Does the following series converge? $$ F(x) = \sum_{n=1}^{\infty} b_n \cos(nx) $$
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The square root function is differentiable in $(0, \pi]$ and so the cosine Fourier series of $ \sqrt x$ converges for x not a multiple of $2 \pi$ . The only problem is at $x = 0$. Think of Dini Test and that the function $\frac{1}{{\sqrt x }}$ is Lebesgue integrable in $[0, \pi]$. This will tell you that the cosine Fourier series too must converge at x = 0. So the convergence of $F(x)$ follows. – Jun 10 '15 at 16:35
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More is true. If you are familiar with function of bounded variation, extending $\sqrt x $ to $[- \pi , \pi] $ as an even function, we get a continuous function of bounded variation in $[ -\pi, \pi]$. Extend to whole of R by periodicity we get a periodic function of period $ 2 \pi$. We can now invoke the well known result that the Fourier series of any periodic continuous function of bounded variation converges uniformly to the function. Therefore, F(x) converges uniformly on $[- \pi, \pi]$ and by periodicity uniformly everywhere. – Jun 11 '15 at 05:07