# Taking derivatives of a power series

I've been working on understanding power series, and came across a problem asking for the derivative of a certain power series and for the derivative to be a summation with a lower limit equal to zero.

Here is the summation:

$$\sum_{k=0}^{\infty} \frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}$$

I tried to write out the sum as terms and then took the derivative to get: $$1-\frac12x^3+\frac7{512}x^6-\frac{10}{110592}x^9+...$$

I am not sure if rewriting this as a summation will give me the answer, but if it will, I do not know how to do this.

• Assuming that $(n!)^3$ is a constant, the summation is $\frac {x}{(n!)^3}\sum_{k=0}^{\infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $\frac {x}{(n!)^3}\frac {1}{1+x^3/2^3}$. – DanielWainfleet Jan 10 '19 at 1:03
• Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything. – Claude Leibovici Jan 10 '19 at 7:10
• Thanks for catching that, it was a typo, and it should be fixed now. – AspiringRoboteer Jan 10 '19 at 13:29

Within the radius of convergence, one can differentiate a power series term by term. Hence, the derivative is $$\frac{d}{dx}\sum_{k=0}^{\infty} \frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}=\sum_{k=0}^{\infty} \frac{(-1)^k (3k+1) x^{3k}}{2^{3k}(k!)^3}$$ This is done by using the power rule for derivatives on each term of the power series.