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I am stuck on this geometric series question:

For what values of $x$ would the infinite geometric series $(2+x)+(2+x)^2+(2+x)^3+\cdots$ converge?

Formula for series:

$$S_n=\frac{a(1-r^n)}{1-r}$$

The first term $a=(2+x)$ and the common ratio $r=(2+x)$ ?

$$S_n=\frac{(2+x)(1-(x+2)^n)}{1-(x+2)}$$

Am I totally wrong on this or somewhat on the right track?

Thanks

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    $\begingroup$ You want $|2 + x| < 1.$ You can solve for $x$ now. You aren't totally wrong, I am guessing you wanted to take the limit from what you have gotten, but I honestly don't find that limit to be very appetizing. $\endgroup$ – IAmNoOne Aug 16 '14 at 23:15
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Yuo're right as far as you've taken it. Now you need to recall the series converges if $-1<r<1$ and diverges otherwise.

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There's an easy formula for geometric series that says it converges if and only if the ratio $r$ has absolute value less than one. In this case you want $-1 < 2 + x < 1$, i.e. $-3 < x < -1$.

However, your way works too. Starting where you left off, we have (for $x \ne -1$) $$ S_n = \frac{(2+x)(1-(x+2)^n)}{1-(x+2)} = \frac{-(2+x)}{1+x}\left(1-(x+2)^n\right) $$ If $|x + 2| < 1$, then $|x+2|^n \to 0$, so the partial sums converge to $\dfrac{-(2+x)}{1+x}$. If $|x + 2| > 1$, $|x + 2|^n \to \infty$ and $(2 + x) \ne 0$, so the partial sums diverge. In the last case $|x + 2| = 1$, we have $x = -1$ or $x = -3$, and you can check that both of these give divergent series. For $x = -1$ your partial sum formula doesn't work, so you need to look at the series directly.

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