Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 6 down vote accepted

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...

share|cite|improve this answer
Ah, I did not notice that! Thank you. – Trent Bing Dec 5 '12 at 23:41

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 $x$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.