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So, I am given a function $f(x) = -\dfrac{1}{(2+x)^2}$ and I am asked the following:

"Compute the Maclaurin series of $f(x)$. Does the series converge? Hint: Use a geometric series."

For the Maclaurin series, I did the following::

$f(x) = \sum\limits_{n=0}^\infty \dfrac{f'^{(n)}(0)}{n!}x^n \\ f^{(n)}(0) = -\frac14, \frac{2*1}8, -\frac{3*2*1}{16}, ... \textrm{ for } n = 0, 1, 2, ... \\ f(x) = \sum\limits_{n=0}^\infty \dfrac{(-1)^{n+1}(n+1)!}{2^{n+2}}\dfrac{x^n}{n!} = \sum\limits_{n=0}^\infty \dfrac{(-1)^{n+1}(n+1)}{4 (2^n)}x^n$

To find the values for which the sequence converges, I did the ratio test: $\lim_{n \to \infty} | \dfrac{(-1)^{n+2}(n+2)x^{n+1}}{4(2^{n+1})}\dfrac{4(2^n)}{(-1)^{n+1}(n+1)x^n} | = \\ |\dfrac{x}{2}| \lim_{n \to \infty} | \dfrac{(n+2)}{(n+1)}| = |\dfrac{x}{2}| < 1 \\ \therefore |x| < 2$

As you can see, however, I did not use geometric series anywhere in that solution. How would I use geometric series to solve this problem?

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You can remark that $\dfrac{d}{dx}\dfrac{1}{2+x} = f(x)$.

Now, the MacLaurin series of $\dfrac{1}{2+x}$ can be easily computed using a geometric series and you should be able to deduce what you want...

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    $\begingroup$ Ah, I did not notice that! Thank you. $\endgroup$ – Brandon Duffany Dec 5 '12 at 23:41
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Start with $$\frac1{2+x}=\frac12\frac1{1-\left(-\frac{x}2\right)}=\frac12\sum_{n\ge 0}\left(-\frac{x}2\right)^n=\frac12\sum_{n\ge 0}\frac{(-1)^n}{2^n}x^n$$

and differentiate it with respect to $x$.

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Apply $z=-x/2$, then differentiate both sides of $$\frac1{1-z}=1+z+z^2+z^3+\ldots$$

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