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:


By solving it we get:


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.

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

2 Answers 2


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 $|x+5|<1$.

  • $\begingroup$ 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. $\endgroup$ Oct 17, 2016 at 6:27
  • $\begingroup$ $$ |x+5|< 1 \iff -1 < x + 5 < 1 \iff -6 < x < -4 $$ $\endgroup$
    – ILoveMath
    Oct 17, 2016 at 6:27
  • $\begingroup$ @ILoveMath I understand that much... I just cannot understand how it ties into the sum of the geometric series. $\endgroup$ Oct 17, 2016 at 6:46
  • $\begingroup$ 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. $\endgroup$ Oct 17, 2016 at 6:52
  • $\begingroup$ @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. $\endgroup$
    – babakks
    Oct 17, 2016 at 7:07

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