<s>[INST] Show power series converges for every $x$. 
Let $$f(x) = 1 + a_{1}x^{1}+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+...$$ be a solution of the differential equation $f'(x)=xf(x).$
Now I need to explain that the power series that define $f(x)$ converges for every $x$.

This problem is introduced by two smaller proofs which I will give (or my attempts for them) so the track of thought is clear.
First I found a simple expression for $\frac{a_{2n}}{a{2n-2}}$ for every $n \in \mathbb{N}$. Since this fraction is $\frac{1}{2}$ when $n=1$, it is $\frac{1}{4}$ when $n=2$, it is $\frac{1}{6}$ when $n=3$ and so on, I let

$$\frac{a_{2n}}{a_{2n-2}}=\frac{1}{2n}.$$

Next I had to show that for every fixed $x \in \mathbb{R}$ there exists a $N \in \mathbb{B}$ such that $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ whenever $n \geq N$. This is my attempt:

Let $x \in \mathbb{R}$ be given and let $N > x^{2}$, then for all $n \geq N : \frac{1}{n} < \frac{1}{x^{2}}.$ Then
$$\frac{|a_{2n}|}{|a_{2n-2}|}=\frac{1}{2} \frac{1}{|n|} < \frac{1}{2} \frac{1}{|x^{2}|}=\frac{1}{2} \left| \frac{x^{2n-2}}{x^{2n}} \right|\\
\Longrightarrow\\
|a_{2n}||x^{2n}| < \frac{1}{2} |x^{2n-2}||a_{2n-2}|.
$$

I expect to have to use $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ to show the convergence, but I have no idea how. Any suggestions?
 [/INST] With $a_{2n}= \frac{1}{2}a_{2n-2}$ you can write your series as the sum of two geometric series.  First take n even.  Define $b_n= a_{2n}$.  Then $b_{n}= \frac{1}{2}b_{n-1}$ so this is a geometric series: $b_0= a_0$, $b_1= a_2= \frac{1}{2}a_0$, $b_2= a_4= \frac{1}{2}a_2= \left(\frac{1}{2}\right)^2a_0$, etc.
And similarly for n odd.  Let $c_n= a_{2n+1}$ so that $c_0= a_1$. $c_1= a_3= \frac{1}{2}a_1$, $c_2= a_5= \frac{1}{2}a_3= \left(\frac{1}{2}\right)^2a_1$ etc.
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