What can we say about the convergence of this series $$\sum {z^n\over n!} $$
I used Alembert's Ration test and get $$\lim_{n \infty}{u_n\over u_{n+1}}={n+1\over z}$$ As this tends to $\infty>1$ can i say that the given series is convergent for all values of $z$ ? 
Note : z is a complex number

You're wrong. The D'Alemebert Ratio test considers
$$
\lim_{n\rightarrow+\infty}\frac{u_{n+1}}{u_n}
$$
NOT
$$
\lim_{n\rightarrow+\infty}\frac{u_{n}}{u_{n+1}}
$$
so correct limit to consider is
$$
\frac{u_{n+1}}{u_n}=\frac{z}{n+1}\stackrel{n\rightarrow+\infty}{\longrightarrow}0\;\;\;\forall z\in\mathbb C
$$
Hence your power series converges on the whole complex plane.
To be precise: the series converges to a well known entire function: actually we have that
$$
\sum_{n=0}^{+\infty}\frac{z^n}{n!}=e^z\;\;\;\forall z\in\mathbb C.
$$
