Consider:
$${1 \over 2}\left[ {{e^{3x}} - {e^{2x}} + {e^x} - 1} \right] = \left( {\sum\limits_{k \ge 0} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {1 \over 2}$$
I understood that because of the expression $1\over 2$, One should define $a_k$ as
$${a_k} = \left\{ {\matrix{ {{{{3^k} - {2^k} + 1} \over {2\cdot k!}},k > 0} \cr {0,\qquad \qquad \,\,\,\, k = 0} \cr } } \right.$$
Can you clarify this for me? I understood that $1\over2$ should be treated as a power-series, but how exactly?
The RHS can be written as $$\begin{align*} \left( {\sum\limits_{k \ge 0} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {1 \over 2}&= \left( {\sum\limits_{k \ge 0} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {1 \over 2}x^0=\\&= \left( {\sum\limits_{k \ge 1} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - {3^0-2^0+1-1 \over 2\cdot0!}x^0=\\&=\left( {\sum\limits_{k \ge 1} {{{{3^k} - {2^k} + 1} \over {2 \cdot k!}}{x^k}} } \right) - 0\cdot x^0 \end{align*}$$ so that the definition of $a_k$ is now straightforward.