A Taylor series about a center is given. Students find the interval of convergence using the ratio test, find the first nonzero terms and general term of the derivative series, show that the derivative series (a geometric series) sums to a given rational function, and decide whether it converges to that function at an endpoint.
A Taylor series about a center is given. Students find the interval of convergence using the ratio test, find the first nonzero terms and general term of the derivative series, show that the derivative series (a geometric series) sums to a given rational function, and decide whether it converges to that function at an endpoint.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
The series is $\displaystyle\sum_{n=1}^{\infty} a_n$ with $a_n=\dfrac{(x-4)^{n+1}}{(n+1)\,3^{n}}$. Apply the ratio test: $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(x-4)^{n+2}}{(n+2)\,3^{n+1}}\cdot\frac{(n+1)\,3^{n}}{(x-4)^{n+1}}\right|=\frac{|x-4|}{3}\cdot\frac{n+1}{n+2}.$$ Taking the limit, $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-4|}{3}\lim_{n\to\infty}\frac{n+1}{n+2}=\frac{|x-4|}{3}.$$ The series converges when $\dfrac{|x-4|}{3}<1$, i.e. $-3<x-4<3$, so $1<x<7$. **Endpoints.** At $x=1$: $x-4=-3$, so $a_n=\dfrac{(-3)^{n+1}}{(n+1)3^n}=\dfrac{(-1)^{n+1}\,3}{n+1}$. This is $\sum \dfrac{(-1)^{n+1}3}{n+1}$, which **converges** by the alternating series test (terms decrease to $0$). At $x=7$: $x-4=3$, so $a_n=\dfrac{3^{n+1}}{(n+1)3^n}=\dfrac{3}{n+1}$. This is $\sum \dfrac{3}{n+1}$, which **diverges** (limit comparison with the harmonic series). **Interval of convergence:** $1\le x<7$.
5 points. P1: sets up a correct ratio a_{n+1}/a_n (with or without absolute value; once shown it is banked against later algebra slips). P2: correct limit |x-4|/3, with no simplification error. P3: interior of the interval, 1<x<7 (or -3<x-4<3 resolved to interval form; |x-4|<3 alone is not enough). P4: considers BOTH endpoints x=1 and x=7. P5: correct analysis at each endpoint (naming alternating series test at x=1 and a comparison/integral test at x=7 suffices) and the correct interval 1<=x<7.
Differentiate the series for $f$ term by term. The general term of $f$ is $\dfrac{(x-4)^{n+1}}{(n+1)3^n}$, so $$\frac{d}{dx}\!\left[\frac{(x-4)^{n+1}}{(n+1)3^n}\right]=\frac{(n+1)(x-4)^{n}}{(n+1)3^n}=\frac{(x-4)^{n}}{3^n}.$$ Thus the Taylor series for $f'$ about $x=4$ has first three nonzero terms and general term $$f'(x)=\frac{x-4}{3}+\frac{(x-4)^2}{9}+\frac{(x-4)^3}{27}+\cdots+\frac{(x-4)^n}{3^n}+\cdots$$
2 points. P6: the first three nonzero terms (x-4)/3, (x-4)^2/9, (x-4)^3/27, listed or written as a polynomial/series. P7: the correct general term (x-4)^n/3^n. The (n+1) factors cancel cleanly, so the differentiated series indexes from n=1 with these terms.
The series for $f'$ is geometric with first term $A=\dfrac{x-4}{3}$ and common ratio $r=\dfrac{x-4}{3}$. On its interval of convergence $|r|<1$, so the geometric sum applies: $$f'(x)=\frac{A}{1-r}=\frac{\dfrac{x-4}{3}}{1-\dfrac{x-4}{3}}=\frac{x-4}{3-(x-4)}=\frac{x-4}{7-x}.$$ This is the required closed form.
1 point. P8 (verification): present the geometric sum ((x-4)/3)/(1-(x-4)/3); this form alone is sufficient to earn the point. The algebra reduces 3-(x-4)=7-x to give (x-4)/(7-x).
No. The radius of convergence of the $f'$ series equals that of the $f$ series, so from part (A) the open interval of convergence of the $f'$ series is $1<x<7$. Since $x=8$ lies **outside** $1<x<7$, the series for $f'$ does not converge there, so it does not converge to $\dfrac{x-4}{7-x}$ at $x=8$. (Equivalently: at $x=8$ the common ratio is $\dfrac{8-4}{3}=\dfrac{4}{3}>1$, so the geometric series $\sum\left(\tfrac{4}{3}\right)^n$ diverges.)
1 point. P9: 'No, x=8 is outside the interval of convergence' is sufficient; can be earned consistently with an incorrect interval imported from part A. The geometric-ratio argument (r=4/3>1 at x=8) is an accepted alternate reason.
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