How do I find the sum of the series $\sum _{ n=1 }^{ \infty }{ (x+5)^n}$ for the values $-6<x<-4$?

Question

Find the values of $x$ for which the series converges.

$$\sum _{ n=1 }^{ \infty }{ (x+5)^{ n } } $$

Find the sum of the series for those values of $x$.

What I did:

I know that I have to have the absolute value of my common ratio, $(x+5)$, be less than $1$, so I set up the following inequality:

$$-1<(x+5)<1$$

By solving it we get:

$$-6<x<-4$$

Now, I am wondering how I can use those values of $x$ to find the sum of the series. I am at a complete loss here, so I would appreciate any hint that will help me solve this on my own.

Answer 1

Use geometric series to get \begin{align} \sum^\infty_{n=1} (x+5)^n = \frac{x+5}{1-(x+5)} = -\frac{x+5}{x+4} \end{align} since $|x+5|<1$.

Answer 2

The sum of the geometric series is \begin{equation*} \sum_{n=0}^{\infty}x^n=\frac{1}{1-x} \end{equation*} if $|x|<1$.

So we have \begin{equation*} \sum_{n=1}^{\infty}(x+5)^n=\sum_{n=0}^{\infty}(x+5)^n-1=\frac{1}{1-(x+5)}-1=-\frac{1}{x+4}-1=-\frac{x+5}{x+4} \end{equation*} if $|x+5|<1$.