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Question
Find the interval of convergence.
Solution
VerifiedAnswered 1 year ago
Answered 1 year ago
Step 1
1 of 3Find 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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