Section 10.14 : Power Series
3. For the following power series determine the interval and radius of convergence.
\[\sum\limits_{n = 0}^\infty {\frac{{n + 1}}{{\left( {2n + 1} \right)!}}{{\left( {x - 2} \right)}^n}} \]Show All Steps Hide All Steps
Start SolutionOkay, let’s start off with the Ratio Test to get our hands on \(L\).
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right){{\left( {x - 2} \right)}^{n + 1}}}}{{\left( {2n + 3} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right){{\left( {x - 2} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 2} \right)\left( {x - 2} \right)}}{{\left( {2n + 3} \right)\left( {2n + 2} \right)\left( {2n + 1} \right)!}}\frac{{\left( {2n + 1} \right)!}}{{\left( {n + 1} \right)}}} \right|\\ & = \left| {x - 2} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{n + 2}}{{\left( {2n + 3} \right)\left( {2n + 2} \right)\left( {n + 1} \right)}} = 0\end{align*}\]Okay, we can see that , in this case, \(L = 0\) for every \(x\).
Show Step 2Therefore, we know that the interval of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty < x < \infty }}\) and the radius of convergence is \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = \infty }}\) .