Note that the bounds on the summations can be omitted since they are implicit in
the binomial coefficient (for non-negative $n$).
Also allow me to use $k$ instead of your $N$ to keep the notation cleaner.
Then we have
$$
\eqalign{
& f(n) = \sum\limits_k {\left( {{1 \over 4}} \right)^{\,2n - k} \sum\limits_l {\left( \matrix{
n \cr
l \cr} \right)\left( \matrix{
n - l \cr
k - 2l \cr} \right)2^{\, - 2l} \;3^{\,n - \left( {k - l} \right)} } } = \cr
& = 3^{\,n} \left( {{1 \over 4}} \right)^{\,2n} \sum\limits_k {\sum\limits_l {\left( \matrix{
n \cr
n - l \cr} \right)\left( \matrix{
n - l \cr
n - \left( {k - l} \right) \cr} \right)\left( {{4 \over 3}} \right)^{\,k - l} \;} } = \cr
& = \left( {{1 \over 4}} \right)^{\,n} \sum\limits_k {\sum\limits_l {\left( \matrix{
n \cr
n - l \cr} \right)\left( \matrix{
n - l \cr
n - \left( {k - l} \right) \cr} \right)\left( {{3 \over 4}} \right)^{\,n - \left( {k - l} \right)} \;} } = \cr
& = \left( {{1 \over 4}} \right)^{\,n} \sum\limits_j {\sum\limits_i {\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
j \cr
i \cr} \right)\left( {{3 \over 4}} \right)^{\,i} \;} } = \cr
& = \left( {{1 \over 4}} \right)^{\,n} \sum\limits_j {\left( \matrix{
n \cr
j \cr} \right)\left( {1 + {3 \over 4}} \right)^{\,j} \;} = \cr
& = \left( {{1 \over 4}} \right)^{\,n} \left( {2 + {3 \over 4}} \right)^{\,n} = \left( {{{11} \over {16}}} \right)^{\,n} \quad \left| {\;0 \le n \in Z} \right. \cr}
$$
Thus the sum $\sum\limits_{n=1}^{\infty} f(n)$ is convergent to $11/5$.