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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.

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4 Answers 4

up vote 3 down vote accepted

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, $a = x+2$.

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I don't know that geometric series, how am I suppose to know it? Is it a transformation of something else? I still don't quite understand, is the only part of the geometric series that I have to memorize the 1-r? –  user138246 Jul 4 '12 at 22:00

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.

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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 $-3 < x < -1$.

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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.

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