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Alternating series remainder

n=0(1)n+1(2n)!\displaystyle\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{(2n)!}} is approximated using the partial sum n=02(1)n+1(2n)!\displaystyle\sum\limits_{n=0}^{2}{\frac{{{\left( -1 \right)}^{n+1}}}{(2n)!}}.
Complete the sentence about the error of the approximation.
The error bound is
and hence the approximation is an
.
What is the absolute value of the error bound?
Round to five decimal places.