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At $x=1$, the left limit is $1$. The right limit is $-1$ because of the period-two extension. Therefore Dirichlet's theorem on Fourier series gives the Fourier series limitEvery Fourier partial sum is also exactly zero there because $\sin ((2j+1)\pi )=0$.

Multiply the complex Fourier series by $e^{-i\omega rt}$ and integrate over one period. Fourier orthogonality removes every term except the one with index $r$, givingFor general square-integrable functions, this is an orthogonal projection onto the corresponding Fourier mode. It remains valid when the expansion converges only in L2 space: integrate finite Fourier partial sums first, then pass to the limit using the Cauchy-Schwarz inequality. No unjustified pointwise exchange of an infinite series and an integral is needed.

A square wave alternates between two constant levels for equal halves of a period. Choose a $2\pi$-periodic function equal to $1$ on $(0,\pi )$ and $-1$ on $(-\pi ,0)$, with value zero at the jumps. Its odd function symmetry makes the mean and every cosine Fourier coefficient vanish. Directly integrating the two constant pieces givesThus only odd sine Fourier harmonics occur, with amplitudes decreasing like the reciprocal of their frequency multiplier. Every finite Fourier partial sum is smooth, yet their pointwise limit has jumps. At those jumps the Fourier series converges to zero, the midpoint of the two levels, regardless of which value we assign to the original function there.

This example explains how smooth components can encode sharp transitions. Increasing the number of Fourier harmonics narrows the transition region, but also produces the persistent overshoot described by the Gibbs phenomenon. The illustrated square wave and coefficient calculation are mathematical constructions, not measurements of a physical device.

For $0<t\le \pi$, split the sum at $\lfloor 1/t\rfloor$. The first part is bounded using $|\sin kt|\le kt$; summation by parts and the geometric bound on interval sine sums control the rest. The harmonic coefficient sum grows like $\log m$, but cancellations keep the whole trigonometric polynomial uniformly bounded. Modulating this polynomial lets a Fourier partial sum expose the large uncancelled half.