Differentiating a convergent power series with "missing terms"

Question

Assume we have $$P(x) = \sum_{k=1}^{\infty} \frac{1}{k} x^{2k}$$

Also assume that it is convergent, even if it isn't. Just serves to illustrate the question. That is every $x$-term with an odd exponent is "missing". Can I still differentiate it like every other power series?

$$P'(x) = \sum_{k=2}^{\infty} \frac{2k}{k} x^{2k-1} = \sum_{k=2}^{\infty} 2 x^{2k-1} = \sum_{k=1}^{\infty} 2 x^{2k+1}$$

Does it also still work if we "eliminate" a few $x$-terms by adding the following for an arbitrary $n \in \mathbb{N}$

$$P_n(x) = \sum_{k=1}^{\infty} \frac{1}{k} x^{2k+n}$$

Answer 1

If you do the substitution $ u = x^2 $, you'll figure it's the same power series. Similarly for the other cases.

Also, you can always differentiate power series term-by-term, because both $f_n$ and $f'_n$ converge uniformly. Wikipedia.

Answer 2

Yes.By using Hadama formula,we know that the convergent radius of this series is $\rho$,such that $\frac{1}{\rho}=\varlimsup_{n \to \infty}\sqrt[\uproot{2} n]{a_n}=\lim_{n \to \infty}\sqrt[\uproot{2} 2n]{\frac{1}{n}}=1$. And in the domain of convergence,we can differentiate term by term.

So you can focus on the radius of convergence and whether specific $x$ lies on that domain,but not care about what if some terms are “missing”.