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?
Best Answer
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...