-7 points details my notes scalcet9 11.9.003.mi.sa. this question has several parts that must be completed…

Explanation:

Step1: Rewrite the function

We rewrite $f(x)=\frac{1}{6 + x}$ as $f(x)=\frac{1}{6(1+\frac{x}{6})}=\frac{1}{6}\cdot\frac{1}{1-(-\frac{x}{6})}$. The geometric - series formula is $\frac{1}{1 - r}=\sum_{n = 0}^{\infty}r^{n}$, where $|r|\lt1$. Here $r =-\frac{x}{6}$.

Step2: Find the power - series representation

Substituting $r =-\frac{x}{6}$ into the geometric - series formula, we get $f(x)=\frac{1}{6}\sum_{n = 0}^{\infty}(-\frac{x}{6})^{n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n + 1}}x^{n}$.

Step3: Determine the interval of convergence

For the geometric series $\sum_{n = 0}^{\infty}r^{n}$ to converge, we need $|r|\lt1$. Since $r=-\frac{x}{6}$, we have $\left|-\frac{x}{6}\right|\lt1$, which simplifies to $|x|\lt6$. When $x = 6$, the series becomes $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n+1}}\cdot6^{n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6}$, which diverges (it is an alternating series whose terms do not tend to 0). When $x=-6$, the series becomes $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n + 1}}\cdot(-6)^{n}=\sum_{n = 0}^{\infty}\frac{1}{6}$, which is a constant - term series and diverges.

Answer:

The power - series representation is $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{6^{n + 1}}x^{n}$, and the interval of convergence is $(-6,6)$.