A Maclaurin series with a known radius of convergence is given. Students determine convergence or divergence at an endpoint, bound the error of a partial sum, find the general term and radius of convergence for the derivative series, and use the ratio test to find the radius of convergence of a related series.
A Maclaurin series with a known radius of convergence is given. Students determine convergence or divergence at an endpoint, bound the error of a partial sum, find the general term and radius of convergence for the derivative series, and use the ratio test to find the radius of convergence of a related series.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
At $x=6$, substitute into $f(x)=\displaystyle\sum_{n=1}^{\infty}\frac{(n+1)x^n}{n^2 6^n}$: $$\sum_{n=1}^{\infty}\frac{(n+1)6^n}{n^2 6^n}=\sum_{n=1}^{\infty}\frac{n+1}{n^2}.$$ Compare with the harmonic series. For every $n\ge 1$, $$\frac{n+1}{n^2}>\frac{n}{n^2}=\frac{1}{n}>0,$$ and $\sum_{n=1}^{\infty}\frac1n$ diverges. By the (direct) comparison test, the series $\sum\frac{n+1}{n^2}$ **diverges**. Therefore the Maclaurin series for $f$ **diverges at $x=6$**.
2 points. 1 point for substituting x=6 to obtain the series sum (n+1)/n^2 (the 6^n cancels); 1 point for a correct divergence conclusion with reason, e.g. comparison with the divergent harmonic series since (n+1)/n^2 > 1/n, or splitting into sum 1/n + sum 1/n^2 where the first diverges. A bare 'diverges' with no justification, or an inconclusive nth-term test (terms -> 0), does not earn the reason point.
At $x=-3$, $f(-3)=\displaystyle\sum_{n=1}^{\infty}\frac{n+1}{n^2}\left(-\tfrac12\right)^n$, an **alternating** series whose term magnitudes $\frac{n+1}{n^2}\left(\frac12\right)^n$ decrease to $0$. By the alternating series error bound, the error in the partial sum $S_3$ is at most the magnitude of the first omitted term (the $n=4$ term): $$|f(-3)-S_3|\le \left|\frac{4+1}{4^2}\left(-\tfrac12\right)^4\right|=\frac{5}{16}\cdot\frac{1}{16}=\frac{5}{256}.$$ Finally compare $\frac{5}{256}$ with $\frac{1}{50}=\frac{5}{250}$. Since $256>250$, $$\frac{5}{256}<\frac{5}{250}=\frac{1}{50}.$$ Therefore $|f(-3)-S_3|\le\frac{5}{256}<\dfrac{1}{50}$, as required.
2 points. 1 point for using the fourth term (n=4) as the error bound, i.e. identifying |a_4| = (5/16)(1/2)^4 = 5/256 from the alternating series error bound; 1 point for the verification that 5/256 < 1/50 (e.g. 1/50 = 5/250 and 256 > 250). The response must justify that the alternating series error bound applies (terms decreasing in magnitude to 0). Using a different term or only stating S3 without the bound does not earn the points.
Differentiate the Maclaurin series for $f$ term by term. The $n$th term is $\dfrac{(n+1)x^n}{n^2 6^n}$, and $$\frac{d}{dx}\!\left[\frac{(n+1)x^n}{n^2 6^n}\right]=\frac{(n+1)\,n\,x^{n-1}}{n^2 6^n}=\frac{(n+1)x^{n-1}}{n\,6^n}.$$ So the general term of the Maclaurin series for $f'$ is $$\boxed{\dfrac{(n+1)x^{n-1}}{n\,6^n}}\quad(n\ge 1).$$ Term-by-term differentiation of a power series does not change the radius of convergence, so the radius of convergence of the series for $f'$ is the same as for $f$: $\mathbf{R=6}$.
2 points. 1 point for the correct general term of f', (n+1)x^{n-1}/(n 6^n) (equivalently (n+1)n x^{n-1}/(n^2 6^n)); 1 point for radius of convergence 6, justified by the fact that differentiating a power series term by term leaves the radius of convergence unchanged. A general term off by the power of x or the n-factor loses the first point.
For $g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{(n+1)x^{2n}}{n^2 3^n}$, apply the ratio test with $a_n=\dfrac{(n+1)x^{2n}}{n^2 3^n}$: $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+2)x^{2n+2}}{(n+1)^2 3^{n+1}}\cdot\frac{n^2 3^n}{(n+1)x^{2n}}=\frac{(n+2)\,n^2}{(n+1)^3}\cdot\frac{x^2}{3}.$$ Taking the limit, $\dfrac{(n+2)n^2}{(n+1)^3}\to 1$, so $$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{x^2}{3}.$$ The series converges when $L<1$: $$\frac{x^2}{3}<1\ \Longrightarrow\ x^2<3\ \Longrightarrow\ |x|<\sqrt{3}.$$ Therefore the radius of convergence of the Maclaurin series for $g$ is $\mathbf{R=\sqrt{3}}$.
3 points. 1 point for setting up the ratio |a_{n+1}/a_n| for the given series; 1 point for the correct limit L = x^2/3 (the polynomial factor in n tends to 1); 1 point for the radius of convergence sqrt(3), obtained from x^2/3 < 1 (so x^2 < 3, |x| < sqrt(3)). Reporting an interval endpoint analysis is not required for radius. An algebra slip in the ratio that changes the base from 3 forfeits the limit and radius points.
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