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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?

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2 Answers

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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 $(*)$

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The series of course depends on where you expand around.

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.

Edit:

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.

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