If $x(t) = \{t\}$ (the fractional part of $t$), then the Fourier series of $x(t)$ is $$\frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\sin(2\pi nt)}{n}$$
My question is why $x(1) = 0.5$ (using the Fourier series) and not $0$, as expected?
Notice that $x(t) = \{t\} = t - \lfloor t \rfloor$, so $x(1) = 1 - \lfloor 1 \rfloor = 1 - 1 = 0$.
However, with the Fourier series, it gives $0.5$ because
$$\begin{split} x(1) &= \frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\sin(2\pi n\cdot1)}{n}\\ &= \frac{1}{2} - \frac{1}{\pi} \cdot 0\\ &= \frac{1}{2} - 0\\ &=\frac{1}{2} \end{split} $$
Is it only true for non-integer values, e.g., is it true for $\sqrt 2$, but not for the number $1$?
The fractional function is continuous everywhere but the integers on the real line. At a continuous point, Fourier series converge nicely to the value of the function, but at a discontinuous point, Fourier series converge to the average of the right and left limits at the point, which gives $\frac{0+1}{2}=0.5$ in your case.
In general, Fourier series is not necessarily convergent. Or Convergence might be in another sense. Pointwise convergence is rare.