The Taylor power-series for a function \(f\) about \(x=0\) converges to \(f\) for \(-1\leq x\leq 1\). The \(n\)th-degree Taylor polynomial for \(f\) about \(x=0\) is given by \(P_n(x)=\displaystyle\sum_{k=1}^n (-1)^k \dfrac{x^k}{k^2+k+1} \). Of the following, which is the smallest number \(M\) for which the alternating power-series error bound guarantees that \(|f(1)-P_4(1)|\leq M\) ?

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\( |f(1)-P_4(x)|\) describes the difference between the actual value versus the value approximated by the Taylor power-series. This error margin will always be less than then \(a_{N+1}\), where \(N\) is the number of terms used in the approximation. In the question, we are using \(P_4(1)\), which is the first four terms of the power-series. Therefore, the absolute value of the remainder will be less than the value of the 5^th term.

$$ M=a_5=\Big|(-1)^5 \frac{(1)^5}{5^2+5+1} \Big| \tag*{at \(x=1, k=5\)}$$ $$ = \boxed{\frac{1}{31}} $$