Differentiation of power series, problem I have the power series 
$$u(x) = \sum_{k=1}^{\infty} \frac{x^{2k+1}}{k(2k+1)} $$
with radius of convergence $R \geq 1 $ and I want to perform termwise derivation for $|x| \lt 1$, but it isn't working for me.
Normally, I would rewrite $u(x)$ so that it is defined for $ k=0,1,2,...$ instead of $k=1,2,3,...$, and then use that 
$$u(x) = \sum_{k=0}^{\infty} a_k x^k \Rightarrow u'(x) = \sum_{k=1}^{\infty} ka_k x^{k-1}$$
but in this case that gives me
$$u(x) = \sum_{k=0}^{\infty} \frac{x^{2k+3}}{(k+1)(2k+3)} \Rightarrow u'(x) = \sum_{k=1}^{\infty} \frac{kx^{2k+2}}{(k+1)(2k+3)}.$$
The answer is supposed to be 
$$u'(x) = \sum_{k=1}^{\infty} \frac{x^{2k}}{k}$$
but as you can see, I'm nowhere near close. What am I doing wrong?
 A: $$u'(x) = \frac{d}{dx}\sum_{k=1}^{\infty} \frac{x^{2k+1}}{k(2k+1)}$$
$$=\sum_{k=1}^{\infty} \frac{d}{dx} \frac{x^{2k+1}}{k(2k+1)}=\sum_{k=1}^{\infty} \frac{(2k+1)x^{2k}}{k(2k+1)}=\sum_{k=1}^{\infty} \frac{x^{2k}}{k}$$


*

*Your formula will not work, keep in mind that
$$\frac{x^{2k+3}}{(k+1)(2k+3)}\ne a_kx^k$$
so you cannot skip the first term, as it won't be equal to $0$.

*You made a mistake when differentiating, note that $$\frac{d}{dx} x^{2k+3}=(2k+3)x^{2k+2}\ne kx^{2k+2}$$
A: Notice that
$$
 \frac{\text{d}}{\text{d}x} \sum_{k=0}^\infty a_k x^\color{red}{k}
 = \sum_{k=1}^\infty a_k \color{red}{k} x^{\color{red}{k}-1}.
$$
The rule is not to just put a $k$ in front, but to pull the exponent down — just like it’s done for polynomials (if you don’t know how to differentiate a polynomial you absolutely need to look this up).
A: You make a mistake in the term-wise differentiation.
$$\frac{d}{dx} \frac{x^{2k+3}}{(k+1)(2k+3)} = \frac{(2k+3) x^{2k+2}}{(k+1)(2k+3)}=\frac{x^{2k+2}}{k+1}$$
Therefore, in your question, it should be
$$u(x) = \sum_{k=0}^{\infty} \frac{x^{2k+3}}{(k+1)(2k+3)} \Rightarrow u'(x) = \sum_{k=0}^{\infty} \frac{x^{2k+2}}{k+1}=\sum_{k=1}^{\infty} \frac{x^{2k}}{k}.$$
