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

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.

• Notices that $x+5$ is between $0$ and $1$ while $-6<x<-4$, this is a geometric series. Oct 17, 2016 at 6:21

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 $|x+5|<1$.
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 $|x+5|<1$.
• How I can utilize the fact that $-6<x<-4$ for $\frac { 1 }{ -x-4 }$? I'm sorry for the stupid question, but this is just elusive for me. Oct 17, 2016 at 6:27
• $$|x+5|< 1 \iff -1 < x + 5 < 1 \iff -6 < x < -4$$ Oct 17, 2016 at 6:27
• We can use the sum of the geometric series only if $|x+5|<1$, that is, only if $-6 < x < -4$, because if $|x+5| \geq 1$ then the series diverges and tends to infinity. Oct 17, 2016 at 6:52
• @Antioquia3943; Sorry, but in your formulation the summation index $n$ has to start from zero, not one. You can verify this by setting $x=0$. Jackie Chong's answer is correct. Oct 17, 2016 at 7:07