Question

Find the interval of convergence.

(1)k2k3k+1xk\sum(-1)^k \frac{2^k}{3^{k+1}} x^k

Solution

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Answered 1 year ago
Answered 1 year ago
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Let the formula part of the given be aka_k. For x0x\neq0, we have

limkak+1ak=limk(1)k+12k+1xx+13k+23k+1(1)k2kxk=limk2x3=2x3\begin{aligned} \lim_{k\to\infty}\left\lvert\dfrac{a_{k+1}}{a_k}\right\rvert=\lim_{k\to\infty}\left\lvert\dfrac{(-1)^{k+1}2^{k+1}x^{x+1}}{3^{k+2}}\cdot\dfrac{3^{k+1}}{(-1)^k2^kx^k}\right\rvert&=\lim_{k\to\infty}\dfrac{2\lvert x\rvert}{3}=\dfrac{2\lvert x\rvert}{3} \end{aligned}

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