Bound for binomial sum Define the function of the positive integers $f(n)$ as follows,
$$ f(n ): = \sum\limits_{N=0}^{2n}  \, \, 
 \, \, 
    \big ( \frac{1}{4}\big )^{2n - N}  
\sum\limits_{\ell=0}^{N/2} 
  \binom{n}{\ell} \, \, 
    \binom{n - \ell}{N - 2 \ell} \,  \, \, 2^{ - 2 \ell}  \, \, 3^{n - ( N - \ell)} $$
Is the sum $\sum\limits_{n=1}^{\infty} f(n)$ convergent?
 A: 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 $11/5$.
