Determinant of a series of Hadamard matrix Given: $$H=\ \left[ \begin{array}{cc|r}
1 & 1  \\ 1 & -1  \end{array} \right]$$
a Hadamard $H_2$ matrix.
and the series:
$$S=\sum_{k=0}^{N}{\frac{H^k}{k!}}$$
Is it possible to calculate the quantity
$D=\det(S)$ 
as a function of N?
Thanks

This is a good example of the power of diagonalization. 
$$H = A \begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix} A^{-1}$$
where it doesn't matter what $A$ is. Then
$$S = A \begin{pmatrix} \sum_{k=0}^N 2^k/k! & 0 \\ 0 & \sum_{k=0}^N (-2)^k/k! \end{pmatrix} A^{-1}$$
so
$$\det S = \left( \sum_{k=0}^N 2^k/k! \right) \left( \sum_{k=0}^N (-2)^k/k! \right).$$
For $N$ large, the first sum approaches $e^2$ and the second sum approaches $e^{-2}$, so the determinant approaches $1$. For a more accurate bound, we have
$$\sum_{k=0}^N 2^k/k! \approx e^2 - 2^{N+1}/(N+1)!$$
and similarly for the other sum, so
$$\left( \sum_{k=0}^N 2^k/k! \right) \left( \sum_{k=0}^N (-2)^k/k! \right) \approx (e^2 - 2^{N+1}/(N+1)!)(e^{-2} - (-2)^{N+1}/(N+1)!)$$
$$\approx 1 - \frac{2^{N+1}}{(N+1)!}(e^{-2} + (-1)^{N+1} e^2) + \mbox{lower order terms}.$$
