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.
Best Answer
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.