Respuesta :
Answer:
The series representation is [tex]\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{6^{n+1}}[/tex] and the interval of convergence is (-6,6)
Step-by-step explanation:
We want to find a series, such that f(x) = \sum_{n=0}{\infty}a_n(x-a)^{n}[/tex], were a is the value that we are using to center the series expansion. In our case, a=0.
We will use the geometric series formula as follows. For |r|<1 then
[tex] \frac{1}{1-r}=\sum_{n=0}^{\infty} r^n[/tex]
In our case, with some algebreaic manipulation we have that
[tex]f(x) = \frac{1}{6+x} = \frac{1}{6}\frac{1}{1-(\frac{-x}{6})}[/tex]
Taking [tex]r = \frac{-x}{6}[/tex] we get that
[tex]f(x) = \frac{1}{6}\sum_{n=0}^{\infty} (\frac{-x}{6})^n = \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{6^{n+1}}[/tex]
This representation is valid (that means that the series converges to the value of f(x)) only for |r|<1. That is
[tex]\left|\frac{-x}{6}\right|= \left|\frac{x}{6}\right|<1[/tex]
which implies that |x|<6. So the interval of convergence is (-6,6).
