Find the radius of convergence and interval of convergence of the series. \sum_{n=3}^{\infty} ...

Question:

Find the radius of convergence and interval of convergence of the series.

{eq}\sum_{n=3}^{\infty} \frac{(-1)^{n-1}x^{n+2}}{n^5} {/eq}

Related Theory:

Radius and Interval of Convergence of any Power Series:

To find the radius and interval of convergence of the power series: {eq}\sum\limits_{n = 0}^\infty {{c_n}{{(x - a)}^n}} {/eq}, we use ratio test in the following manner:

Step 1: Let {eq}{a_n} = {c_n}{(x - a)^n} {/eq} and {eq}{a_{n + 1}} = {c_{n + 1}}{(x - a)^{n + 1}} {/eq}

Step 2: Simplify the ratio: {eq}\left| {\frac{{a{}_{n + 1}}}{{{a_n}}}} \right| = \left| {\frac{{{c_{n + 1}}{{(x - a)}^{n + 1}}}}{{{c_n}{{(x - a)}^n}}}} \right| = \left| {\frac{{{c_{n + 1}}}}{{{c_n}}}(x - a)} \right| {/eq}

Step 3: Find the interval and radius of convergence. The series is convergent whenever {eq}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a{}_{n + 1}}}{{{a_n}}}} \right| < 1 {/eq}

In this case: {eq}\displaystyle \sum_{n=3}^{\infty} \frac{(-1)^{n-1}x^{n+2}}{n^5} {/eq}

Step 1: here, {eq}{a_n} = \displaystyle...

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