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.
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.
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, $a = x+2$.
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
So you'd need $-1
That's the same as $-3
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
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.