Series convergence, finding values that cause convergence. I am trying to find the $x$ values that make this series converge:
$$\sum_{n = 1}^\infty (x+2)^n.$$
To me it seems like $x = -2$ would make the series converge but that is a wrong answer, I am not sure why either.
 A: Recall the geometric progression $$\sum_{n=1}^{N} a^n = a \left(\dfrac{1-a^{N}}{1-a} \right)$$ and hence the geometric series $\displaystyle \sum_{n=1}^{\infty} a^n$ converges if and only if $\vert a \vert < 1$.
In your problem, $a = x+2$.
A: If you're familiar with the infinite geometric series
$$
\sum_{n=0}^\infty r^n = 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r}
$$
then the series you've got is the same thing with $r=x+2$.
The series above converges if $-1<r<1$ and otherwise diverges.
So you'd need $-1<x+2<1$.
That's the same as $-3<x<-1$.  So that's the interval of convergence.
A: The well known truncated (i.e. not infinite) geometric series is
$$S=\sum_{n=0}^k a^n=1+a+a^2+\cdots+a^k$$
which can be written as $$\frac{1-a^{k+1}}{1-a}$$ 
because, if we multiply the entire sum by $a$, we get
$$Sa=a+a^2+\cdots+a^{k+1}$$  
and taking the difference between the original sum and the sum multiplied by $a$, we solve for $S$:
$$S-Sa=S(1-a)=1+a+a^2+\cdots+a^k-(a+a^2+\cdots+a^{k+1})=1-a^{k+1} \implies\\ S=\frac{1-a^{k+1}}{1-a}$$
Now, we see that, setting $k \to \infty$
$$\sum_{n=0}^\infty a^n=\lim_{k \to \infty}\sum_{n=0}^k=\lim_{k \to \infty}\frac{1-a^{k+1}}{1-a}$$
only converges when $|a|<1$, because otherwise $a^{k+1}$ (in the numerator) would become infinitely large!
Thus, if we substitute $a=x+2$ into the above equation, we get
$$\sum_{n=0}^\infty (x+2)^n=\lim_{k \to \infty}\frac{1-(x+2)^{k+1}}{1-(x+2)}$$
which, as we determined, converges only for $|a|=|x+2|<1$.  This equivalence can be rewritten as $-1<x+2<1$, and, by subtracting $2$ from everything, we get $-3 < x < -1$.  
A: You are to find all values of $x$ for which the series converges. $x=-2$ is just one of them. Then answer will be in the form of an interval, the interval of convergence.
