What is the relationship between Fourier transformation and Fourier series? Is there any connection between Fourier transformation of a function and its Fourier series of the function? I only know the formula to find Fourier transformation and to find Fourier coefficients to find the corresponding Fourier series.

One can (at least on an informal level) think of Fourier transforms as a kind of limit of Fourier series.
If $f$ is a function on $\mathbb R$, we can restrict it to the interval $[N,-N]$,
where it has a Fourier series, namely
$$f(x) = \sum_{n = -\infty}^{\infty} a_n  e^{\frac{\pi i n x}{N}} ,$$
where $$a_n = \frac{1}{2N}\int_{-N}^N f(x) e^{\frac{-\pi i n x}{N}}dx
= \frac{1}{2N}\int_{-N}^N  f(x)e^{-ix y} dx,$$ where $y = \pi n/N$.
This is valid for $x \in [-N,N].$
So $$f(x) = \sum_{n = - \infty}^{\infty} e^{ixy} \cdot \int_{-N}^N
f(x) e^{-ixy} dx \cdot \frac{\Delta y}{2\pi},$$ where $y = \pi n/N$, and $\Delta y = \pi/N$
(again, valid for $x \in [-N,N]$).
Letting $N \to \infty$, defining $$\hat{f}(y) = \int_{-\infty}^{\infty}
f(x)e^{-ixy} dy,$$ and thinking of the sum as a Riemann sum, we find that
$$f(x) = \frac{1}{2\pi}  
\int_{-\infty}^{\infty} \hat{f}(y)e^{ixy} dy,$$
now valid for all $x \in (-\infty,\infty).$
This is the representation of $f$ as the integral of its Fourier transform.
(The above is informal, since I didn't carefully discuss convergence issues.  It is a very traditional and well-known intuition, though, and can be made rigorous in various contexts; e.g. my memory is that Wiener proved his Tauberian theorem by passing from the Fourier series to the Fourier transform context using this sort of argument.)
