# use differentiation to find a power series representation for 1/(3+x)^2

would anyone tell me how to solve this?

Use differentiation to find a power series representation for

$$\frac{1}{(3+x)^{2}}$$

What is the radius of convergence, R?

• Expanding around x = 0? – Ben Lansdell Nov 21 '16 at 6:26
• @ lansdell yes. I think so. – Tmm Nov 21 '16 at 6:28
• What is the derivative of $-\dfrac{1}{x+3}$? Can you find the power series representation of $-\dfrac{1}{x+3}$? – John Wayland Bales Nov 21 '16 at 6:28

Possible hint: We have $$1/(3 + x)=(1/3)\frac{1}{(1-(-x/3))}$$ Use this fact that $1/(1-t)=\sum_0^{\infty}t^n~~~(*)$ where $|t|<1$. I mean take $t=-x/3$ and...
Note that $|-t|=|t|<1$ is the radius of convergence of $(*)$
Taking $x = 0$ as an example: $$f(x) = \frac{1}{(x+3)^2} = \sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!},$$ where $$f^{(n)}(x) = \frac{(-1)^n(n+1)!}{(x+3)^{n+2}}.$$ So the series is $$\frac{1}{(x+3)^2} = \sum_{n=0}^\infty \frac{(-1)^n(n+1)!}{(3)^{n+2}}\frac{x^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n(n+1)}{(3)^{n+2}}x^n,$$ This would have a radius of convergence of 3.
As the problem stipulates you should use differentiation, note that another way to get to the same series, as hinted at, is by noting: $$\frac{1}{(x+3)^2} = -\frac{d}{dx}\frac{1}{(x+3)}$$ where $$\frac{1}{(x+3)} = \frac{1}{3}\frac{1}{1-(-x/3)} =\frac{1}{3}\sum^\infty_{n=0}[-x/3]^n$$ provided $|-x/3|<1$ ie provided $|x|<3$. Thus differentiating this series gives you the same answer as above.