I need to find the laurent series and the residue of the following complex function
$$f(z)=(z+1)^2e^{3/z^2}$$
at $z=0$.

Since $e^z=\sum z^n/n!$, then 
$$e^{3/z^2}=\sum\_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$
thus
$$f(z)=(z^2+2z+1)\sum\_{n=0}^\infty \frac{3^n/n!}{z^{2n}}=\sum\_{n=0}^\infty \frac{3^n/n!}{z^{2(n-1)}}+\sum\_{n=0}^\infty \frac{2\cdot3^n/n!}{z^{2n-1}}+\sum\_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$
which, with a shift of index and expansion of positive powers, can be expressed as
$$f(z)=z^2+3+\sum\_{n=1}^\infty\left(\frac{3}{n+1}+1\right)\frac{3^n/n!}{z^{2n}}+\sum\_{n=1}^\infty\frac{2\cdot3^n/n!}{z^{2n-1}}$$
so the residue is given by evaluating the numerator of the second series at $n=1$, so its value is $6$. I tried using WolframAlpha and Mathematica to check my answer, but both would not return a value. Would this be correct? Also, is there a way to put the two sums together (one gives the coefficients of even powers, while the other of the odd) so I can have the principal part of the laurent series expressed only with one sum?