Prove that the series is convergent and find the sum of it.
$$ \sum_{n=0}^{\infty} \left( \sum_{k=0}^n (-2) ^{-k}3 ^{-n+k} \frac{1}{k!} \right) $$
I've tried to do something with calculating this sum with respect to k but without any satisfying result.
Once you have congergence (see Antonio Vargas' comment), you can proceed by exchanging the order of summation:
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-2)^{-k} 3^{-n+k}}{k!} = \sum_{k=0}^\infty \sum_{n=k}^\infty \frac{(-2)^{-k} 3^{-n+k}}{k!} = \sum_{k=0}^\infty \frac{(-2)^{-k} 3^k}{k!} \sum_{n=k}^\infty 3^{-n}$$
The inner sum now is $\frac{1}{1 - 3^{-1}} - \frac{1 - 3^{-k}}{1 - 3^{-1}} = \frac{3^{-k}}{1 - 3^{-1}} = \frac{3}{2} 3^{-k}$, so we get $$\frac{3}{2} \sum_{k=0}^\infty \frac{(-2)^{-k}}{k!} = \frac{3}{2} \exp(-1/2)$$