find the function represented by the following series and find the interval of convergence of the series…

find the function represented by the following series and find the interval of convergence of the series. \n∑_{k = 0}^∞ ((x^2 + 2)/6)^k\nthe function represented by the series ∑_{k = 0}^∞ ((x^2 + 2)/6)^k is f(x)=□.

find the function represented by the following series and find the interval of convergence of the series. \n∑_{k = 0}^∞ ((x^2 + 2)/6)^k\nthe function represented by the series ∑_{k = 0}^∞ ((x^2 + 2)/6)^k is f(x)=□.

Answer

Explanation:

Step1: Recognize geometric - series form

A geometric series has the form $\sum_{k = 0}^{\infty}r^{k}=\frac{1}{1 - r}$, where $|r|\lt1$. In the given series $\sum_{k = 0}^{\infty}(\frac{x^{2}+2}{6})^{k}$, we have $r=\frac{x^{2}+2}{6}$.

Step2: Find the function

Using the formula for the sum of an infinite geometric series, $f(x)=\frac{1}{1-\frac{x^{2}+2}{6}}$. Simplify the expression: [ \begin{align*} f(x)&=\frac{1}{\frac{6-(x^{2}+2)}{6}}\ &=\frac{6}{6 - x^{2}-2}\ &=\frac{6}{4 - x^{2}} \end{align*} ]

Step3: Find the interval of convergence

For a geometric series $\sum_{k = 0}^{\infty}r^{k}$ to converge, $|r|\lt1$. So we set $\left|\frac{x^{2}+2}{6}\right|\lt1$. Since $x^{2}+2\gt0$ for all real - valued $x$, we have $\frac{x^{2}+2}{6}\lt1$. [ \begin{align*} x^{2}+2&\lt6\ x^{2}&\lt4\

  • 2\lt x&\lt2 \end{align*} ]

Answer:

The function is $f(x)=\frac{6}{4 - x^{2}}$, and the interval of convergence is $(-2,2)$.