Calculate the Taylor-series for $f : x \mapsto \frac{1}{1+2x}$ and find its radius of convergence

Question

From an old exam:

Calculate the Taylor-series for $f : x \mapsto \frac{1}{1+2x}$ at $x_{0} = 0$ . What's the radius of convergence of it?

Alright, so it wasn't mentioned how many derivates we gonna need, I guess the more the better / clear it is. Here I have created the first 4 derivatives:

$$f^{(0)} = \frac{1}{1+2x}$$

$$f^{(1)}= -\frac{2}{\left ({1+2x}\right )^{2}}$$

$$f^{(2)}= \frac{8}{\left ({1+2x}\right )^{3}}$$

$$f^{(3)}= -\frac{48}{\left ({1+2x}\right )^{4}}$$

$$f^{(4)}= \frac{384}{\left ({1+2x}\right )^{5}}$$

$$\Rightarrow$$

$$\left ( T_{4,0}f \right ) (x) = \frac{f^{(0)}(0)}{0!} \cdot \left (x-0 \right )^{0} + \frac{f^{(1)}(0)}{1!} \cdot \left (x-0 \right )^{1} + \frac{f^{(2)}(0)}{2!} \cdot \left (x-0 \right )^{2} + \frac{f^{(3)}(0)}{3!} \cdot \left (x-0 \right )^{3} + \frac{f^{(4)}(0)}{4!} \cdot \left (x-0 \right )^{4} + ...$$

$$= \frac{1}{1}\cdot 1 + \frac{-2}{1} \cdot x + \frac{8}{2} \cdot x^{2} + \frac{-48}{6} \cdot x^{3} + \frac{384}{24} \cdot x^{4} + ...$$

$$= 1-2x+4x^{2}-8x^{3}+16x^{4}-...$$

From this, I made the series:

$$\sum_{n=0}^{\infty} \left( -1 \right)^{n} \cdot 2^{n}\cdot x^{n}$$

I believe I did it right so far...?

And now the radius of convergence humm... For this we have always used the ratio test (ignore the $x^{n})$ and just do ratio test. In the end, take the reciprocal of the result and it will be your radius of convergence.

But how would you use ratio test on the series above..? It seems impossible.

Answer 1

What you did above is correct. However, it doesn't candidate for a formal proof, as you just "intuited" the higher derivatives, without actually computing them. You have two possibilities:

1. Prove by induction that $f^{(n)}(0) = (-2)^n$. Do it, it's a good exercise. Then it is easy co compute the radius of convergence via e.g. the ratio test.
2. The smart way to do it: we know that $\frac{1}{1-y} = \sum_{n\ge0}y^n$ with radius of convergence $1$. Now your function is $$\frac{1}{1-(-2x)} = \sum_{n\ge0}(-2x)^n = \sum_{n\ge0}(-2)^nx^n$$ with radius of convergence determined by $|-2x|<1$, that is $x<\tfrac{1}{2}$.