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

Explanation:

Step1: Recognize geometric - series form

A geometric series has the form $\sum_{k = 0}^{\infty}ar^{k}$, where $a$ is the first - term and $r$ is the common ratio. In the given series $\sum_{k = 0}^{\infty}5e^{-6kx}=5\sum_{k = 0}^{\infty}(e^{-6x})^{k}$, we have $a = 5$ and $r=e^{-6x}$.

Step2: Use the formula for the sum of an infinite geometric series

The sum of an infinite geometric series $\sum_{k = 0}^{\infty}ar^{k}=\frac{a}{1 - r}$ when $|r|\lt1$. Substituting $a = 5$ and $r = e^{-6x}$ into the formula, we get $f(x)=\frac{5}{1 - e^{-6x}}=\frac{5e^{6x}}{e^{6x}-1}$ for $|e^{-6x}|\lt1$.

Step3: Find the interval of convergence

We solve the inequality $|e^{-6x}|\lt1$. Since $e^{-6x}=\frac{1}{e^{6x}}\gt0$ for all real $x$, the inequality $e^{-6x}\lt1$ is equivalent to $\frac{1}{e^{6x}}\lt1$, or $e^{6x}\gt1$. Taking the natural logarithm of both sides, $\ln(e^{6x})\gt\ln(1)$. Since $\ln(e^{6x}) = 6x$ and $\ln(1)=0$, we have $6x\gt0$, so $x\gt0$.

Answer:

The series represents the function $f(x)=\frac{5e^{6x}}{e^{6x}-1}$, and the interval of convergence is $(0,\infty)$