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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?

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    $\begingroup$ You should explain why this sum is of interest, or at least what you tried $\endgroup$ – charmd Feb 16 '18 at 16:17
  • $\begingroup$ The binomial coefficient in the second sum will be 0 for $l>N/2$. Why do you sum the zeros up to $l=N$? $\endgroup$ – user Feb 16 '18 at 16:25
  • $\begingroup$ This is just a sum that happens to be relevant for what I am doing I don't understand whether it is infinite or finite, so there is a challenge and this is already a good motivation for doing mathematics. I am trying to modify the terms of the sum in order to obtain something that might look familiar, but I fail. So I decided to share this problem with the community. $\endgroup$ – QuantumLogarithm Feb 16 '18 at 16:26
  • $\begingroup$ You are right, I can reduce the sum to $\ell < N/2$ (see edit). $\endgroup$ – QuantumLogarithm Feb 16 '18 at 16:27
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    $\begingroup$ In any case you can cancel $(1/2)^N$ and $2^N$. $\endgroup$ – user Feb 16 '18 at 16:33
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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 $11/5$.

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