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Question

# Find the interval of convergence.$\sum \frac{k}{2 k+1} x^{2 k+1}$

Solution

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Find the interval of convergence using thw Ratio Test. Calculate $\lim\limits_{k\rightarrow\infin}\left| \dfrac{b_{k+1}}{b_k} \right|$. \begin{align*} \lim\limits_{k\rightarrow\infin}\left| \dfrac{b_{k+1}}{b_k} \right|&= \lim\limits_{k\rightarrow\infin}\left| \dfrac{\dfrac{k+1}{2k+3}x^{2k+3}}{\dfrac{k}{2k+1}x^{2k+1}} \right| \\ \end{align*} Cancel like factors. \begin{align*} \lim\limits_{k\rightarrow\infin}\left| \dfrac{\dfrac{k+1}{2k+3}\cancel{x^{2k+1}}x^2}{\dfrac{k}{2k+1}\cancel{x^{2k+1}}} \right| &= x^2\lim\limits_{k\rightarrow\infin} \dfrac{(k+1)(2k+1)/:k^2}{k(2k+3)/:k^2}\\ &=x^2\lim\limits_{k\rightarrow\infin} \dfrac{(1+\frac{1}{k})(2+\frac{1}{k})}{2+\frac{3}{k}}\\ &=x^2 \end{align*} Solve inequality $x^2<1$. \begin{align*} x^2&<1\\ -1<&x<1 \end{align*} This power series converge for $x\in\langle-1,1\rangle$.

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