complete parts a through c for the function below. f(x)=5 sin 3x a. find the first four nonzero terms of the…

Explanation:

Step1: Recall ratio - test formula

The ratio - test for a power series $\sum_{k = 0}^{\infty}a_{k}x^{k}$ uses $L=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}x^{k+1}}{a_{k}x^{k}}\right|=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|\cdot|x|$. For the power series $5\sin(3x)=\sum_{k = 0}^{\infty}\frac{5(-1)^{k}3^{2k + 1}}{(2k+1)!}x^{2k + 1}$, we have $a_{k}=\frac{5(-1)^{k}3^{2k + 1}}{(2k+1)!}$.

Step2: Calculate $\frac{a_{k + 1}}{a_{k}}$

$a_{k+1}=\frac{5(-1)^{k + 1}3^{2(k + 1)+1}}{(2(k + 1)+1)!}=\frac{5(-1)^{k + 1}3^{2k+3}}{(2k + 3)!}$. Then $\frac{a_{k + 1}}{a_{k}}=\frac{\frac{5(-1)^{k + 1}3^{2k+3}}{(2k + 3)!}}{\frac{5(-1)^{k}3^{2k + 1}}{(2k+1)!}}$. Simplify this expression: [ \begin{align*} \frac{a_{k + 1}}{a_{k}}&=\frac{5(-1)^{k + 1}3^{2k+3}(2k + 1)!}{5(-1)^{k}3^{2k + 1}(2k + 3)!}\ &=\frac{(-1)\times3^{2}}{(2k+2)(2k + 3)}\ &=\frac{-9}{(2k+2)(2k + 3)} \end{align*} ]

Step3: Find the limit as $k\rightarrow\infty$

$L=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|\cdot|x|=\lim_{k\rightarrow\infty}\left|\frac{-9}{(2k+2)(2k + 3)}\right|\cdot|x|$. Since $\lim_{k\rightarrow\infty}\frac{9}{(2k+2)(2k + 3)} = 0$, then $L = 0\cdot|x|=0$ for all $x\in R$.

Step4: Determine the interval of convergence

Since $L<1$ for all $x\in R$, the interval of convergence of the power - series $\sum_{k = 0}^{\infty}\frac{5(-1)^{k}3^{2k + 1}}{(2k+1)!}x^{2k + 1}$ is $(-\infty,\infty)$.

Answer:

$(-\infty,\infty)$