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

Explanation:

Step1: Rewrite the series

We can rewrite the series $\sum_{k = 1}^{\infty}\frac{(x - 2)^k}{3^{2k}}$ as $\sum_{k = 1}^{\infty}\frac{(x - 2)^k}{9^{k}}=\sum_{k = 1}^{\infty}(\frac{x - 2}{9})^k$. This is a geometric - series with common ratio $r=\frac{x - 2}{9}$. The general form of a geometric series is $\sum_{k = 0}^{\infty}ar^k$, and our series starts from $k = 1$. The sum of an infinite geometric series $\sum_{k = 0}^{\infty}ar^k=\frac{a}{1 - r}$ for $|r|\lt1$. Our series $\sum_{k = 1}^{\infty}(\frac{x - 2}{9})^k$ can be obtained from the standard geometric series. The sum of the geometric series $\sum_{k = 0}^{\infty}r^k=\frac{1}{1 - r}$ for $|r|\lt1$. So, $\sum_{k = 1}^{\infty}r^k=\sum_{k = 0}^{\infty}r^k-1=\frac{1}{1 - r}-1=\frac{r}{1 - r}$. Here $r=\frac{x - 2}{9}$, so the function $f(x)=\frac{\frac{x - 2}{9}}{1-\frac{x - 2}{9}}$.

Step2: Simplify the function

Simplify $f(x)=\frac{\frac{x - 2}{9}}{1-\frac{x - 2}{9}}=\frac{x - 2}{9-(x - 2)}=\frac{x - 2}{11 - x}$, for $|\frac{x - 2}{9}|\lt1$.

Step3: Find the interval of convergence

Solve the inequality $|\frac{x - 2}{9}|\lt1$. We have $- 1\lt\frac{x - 2}{9}\lt1$. Multiply each part of the compound - inequality by 9: $-9\lt x - 2\lt9$. Add 2 to each part: $-9 + 2\lt x\lt9+2$, so $-7\lt x\lt11$. We need to check the endpoints. When $x=-7$, the series becomes $\sum_{k = 1}^{\infty}(\frac{-7 - 2}{9})^k=\sum_{k = 1}^{\infty}(-1)^k$, which diverges (divergent geometric series with $|r| = 1$ and non - zero terms). When $x = 11$, the series becomes $\sum_{k = 1}^{\infty}(\frac{11 - 2}{9})^k=\sum_{k = 1}^{\infty}1^k$, which diverges.

Answer:

The function is $f(x)=\frac{x - 2}{11 - x}$, and the interval of convergence is $(-7,11)$.