What is the difference between Fourier series and Fourier transformation? What's the difference between Fourier transformations and Fourier Series? 
Are they the same, where a transformation is just used when its applied (i.e. not used in pure mathematics)?
 A: Fourier transform and Fourier series are two manifestations of a similar idea, namely, to write general functions as "superpositions" (whether integrals or sums) of some special class of functions. Exponentials $x\rightarrow e^{itx}$ (or, equivalently, expressing the same thing in sines and cosines via Euler's identity $e^{iy}=\cos y+i\sin y$) have the virtue that they are eigenfunctions for differentiation, that is, differentiation just multiplies them: ${d\over dx}e^{itx}=it\cdot e^{itx}$. This makes exponentials very convenient for solving differential equations, for example.
A periodic function provably can be expressed as a "discrete" superposition of exponentials, that is, a sum. A non-periodic, but decaying, function does not admit an expression as discrete superposition of exponentials, but only a continuous superposition, namely, the integral that shows up in Fourier inversion for Fourier transforms.
In both case, there are several technical points that must be addressed, somewhat different in the two situations, but the issues are very similar in spirit.
A: Fourier transform is used to transform periodic and non-periodic signals from time domain to frequency domain. It can also transform Fourier series into the frequency domain, as Fourier series is nothing but a simplified form of time domain periodic function.
Fourier series


*

*Periodic function => converts into a discrete exponential or sine and cosine function. 

*Non-periodic function => not applicable
Fourier transform


*

*Periodic function => converts its Fourier series in the frequency domain. 

*non-Periodic function => converts it into continuous frequency domain.
A: If you have a locally compact Abelian group $G$ you can define a group called the Pontryagin dual group - $\widehat{G}$. You can define a Haar measure on $G$, $\mu$. We can define the Fourier transform of a function $f\in L^1(G)$:
$$\widehat f(\chi)=\int_Gf(x)\overline{\chi(x)}d\mu(x)$$
$\widehat f(\chi)$ is a bounded continuous function that vanishes at infinity on $\widehat{G}$.
If $G=\Bbb R$ then $\widehat{G}=\Bbb R$ and we have the regular Fourier transform.
If $G=S^1$ then $\widehat{G}=\Bbb Z$ and we have the Fourier series (an example of a Fourier transform).
A: The Fourier series is used to represent a periodic function by a discrete sum of complex exponentials, while the Fourier transform is then used to represent a general, nonperiodic function by a continuous superposition or integral of complex exponentials. The Fourier transform can be viewed as the limit of the Fourier series of a function with the period approaches to infinity, so the limits of integration change from one period to $(-\infty,\infty)$.
In a classical approach it would not be possible to use the Fourier transform for a periodic function which cannot be in $\mathbb{L}_1(-\infty,\infty)$. The use of generalized functions, however, frees us of that restriction and makes it possible to look at the Fourier transform of a periodic function. It can be shown that the Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function.
A: I see you’re asking probably about the most “meaningful two ones” that are seemingly differ, but yet comparable if you compare them one to one and these are discrete time signal’s fourier transform and discrete time signal’s fourier series, while both signals are periodic. 
I’ll explain it in the simplest way possible. 
Imagine you have a drawing in a map (don’t imagine an nth dimensional map, but it might be). Both of them represent a point on this drawing on this map. (Map is a globe map, don’t go too far.) Now it’s simple, each representation solves a different problem, but both are representing the same point. It’s just one representation represent a point as a point as we know it  (alas series, with e and i or j and w0) and one representation (the one with the deltas) represents the same point as  + <0,delta(y)>. We know the first representation. I won’t explain. But in the second representation if we sum up all e-s with i-s or j-s we’ll get what? A superposition. Meaning that we’ll know the final superposition, but won’t know what the projection is. We won’t know what x,y is because in signal x,y are not trivial to see as a point on the map while they are in superposition, so we need a coordinate system instead. The Fourier Transform shows us the same point, but like in the map - with coordinates. Alas we see the coefficients. And we see the coefficients thanks to the deltas. Otherwise we’d had a sum of e’s which is sort of “summable” sometimes and in order to find each coefficient we’d had to project and calculate the projection for each coefficient. Here we SEE the coefficients WITHOUT calculating each of them thanks to the deltas. 
All this is because the inner product in signal’s dimension is “less straight forward” than the inner product in a globe map. But yet it is straight forward for a computer if you do it repeatedly and ineffectively. :)
That’s it, hope it helps. 
