Trying to solve a Taylor series problem I have a Taylor series problem, well more precisely a Maclaurin series.
I am trying to find convergence of: $f(x) = e^{x^3} + e^{{2x}^3}$
Okay here goes:
$$f'(x) = 3xe^{x^3} + 6x e^{{2x}^3}$$
$$f''(x) = 9x^2e^{x^3} + 3e^{x^3} + 36x^2e^{{2x}^3} + 6e^{{2x}^3}=e^{x^3}(9x^2+3) + e^{{2x}^3}(36x^2+6)$$
$$f'''(x) = (9x^2+3)(3xe^{x^3}) + 18xe^{x^3} + (36x^2 + 6)(6xe^{2x^3}) + 72xe^{2x^3}$$
$$=27xe^{x^3}(x^2 + 1) + 108xe^{2{x^3}}(2x^2 + 1)$$
Now I don't see a pattern, but I can note at $a=0$ we have:
$$f(0)=2$$
$$f'(0)=0$$
$$f''(0)=9$$
$$f'''(0)=0$$
I am not sure where this is going, or if this is the right way to attack the problem. Any advice?
 A: Use the familiar Maclaurin expansion for the exponential:
$$e^t=1+t+\frac{t^2}{2!}+\dots=\sum_{n=0}^\infty \frac{t^n}{n!}. $$
If you plug in $t=x^3$ you get
$$e^{x^3}=\sum_{n=0}^\infty \frac{x^{3n}}{n!}. $$
If you plug in $t=2x^3$ you similarly get
$$e^{2x^3}=\sum_{n=0}^\infty 2^n \frac{x^{3n}}{n!} $$
Try adding these two expansions.
A: To supplement the other answer, here's how you would determine convergence using the ratio test:
We would like to test the convergence of the sum
$$
\sum_{n=1}^\infty a_{3n} x^{3n} = \sum_{n=1}^\infty \frac{1 + 2^n}{n!}x^{3n}
$$
In order to do so, we need to determine the value of
$$
\limsup_{n \to \infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = 
\lim_{n \to \infty} 
\frac{(1 + 2^{n+1})x^{3(n+1)}}{(n+1)!} \cdot
\frac{n!}{(1 + 2^n)x^{3n}}
$$
Simpifying, this becomes
$$
\lim_{n \to \infty} \frac{x^3}{n+1} \cdot \frac{1+2^{n+1}}{1+2^n}
$$
Noting that this limit becomes zero for all $x$ and noting that $0<1$, we conclude that this sum converges for all $x$ by the ratio test.
