How condition for existence of Fourier transform is valid?
As mentioned in Batman's answer, the condition of the sequence being absolutely summable is only sufficient but not necessary. The Fourier transform can be extended to $\ell_2$ sequences, i.e. sequences for which

$$\sum_{n=-\infty}^{\infty}|f[n]|^2&lt;\infty$$

is satisfied. A further generalization is possible if you allow distributions and their derivatives (such as the Dirac delta function) as Fourier transforms. Then also sequences with polynomial growth have a Fourier transform in that generalized sense. Just sequences with exponential growth have no Fourier transform. However, they can be handled by the $\mathcal{Z}$-transform. Note that everything said above also holds for continuous-time functions (if your replace "$\mathcal{Z}$-transform" by "Laplace transform").

Sequences with a constant amplitude (such as the step function or a sinusoid) will have a Dirac delta in their Fourier transform. Sequences with a linear growth will have the (generalized) derivative of a Dirac delta in their Fourier transform, etc.

And as for your question how it is possible for a sum over infinitely many elements to be finite, your should as a first basic example read up on the geometric series.

EDIT:

So if we allow a Dirac delta impulse as a Fourier transform, we can ask what the corresponding time function would be. By taking the inverse Fourier transform of a shifted Dirac delta impulse $\delta(\omega-\omega_0)$ we get

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega-\omega_0)e^{j\omega t}d\omega=\frac{1}{2\pi}e^{j\omega_0t}$$

which follows from the sifting property of the Dirac delta impulse. So we now have the Fourier transform pair

$$e^{j\omega_0t}\Longleftrightarrow2\pi\delta(\omega-\omega_0)$$

from which the transforms of $\cos(\omega_0t)$ and $\sin(\omega_0t)$ follow directly by rewriting these functions as $\cos(\omega_0t)=\frac12(e^{j\omega_0t}+e^{-j\omega_0t})$ and $\sin(\omega_0t)=\frac{1}{2j}(e^{j\omega_0t}-e^{-j\omega_0t})$.