Yes, any series that is obtained through differentiation or integration has the same radius of convergence as the series it came from.
The only place it might be different is at the endpoints. For example, $\displaystyle \begin{align*} \frac{1}{1 - z} = \sum_{n = 0}^{\infty}{ z^n } \end{align*}$ is convergent for $\displaystyle \begin{align*} |z| < 1 \end{align*}$ and divergent for $\displaystyle \begin{align*} |z| \geq 1 \end{align*}$, while if we integrate it we have
$\displaystyle \begin{align*} \frac{1}{1 - z} &= \sum_{n = 0}^{\infty}{z^n} \\ \int{ \frac{1}{1 - z}\,\mathrm{d}z } &= \int{ \sum_{n = 0}^{\infty}{z^n}\,\mathrm{d}z } \\ -\ln{ \left( 1 - z \right) } &= \sum_{n = 0}^{\infty} {\frac{z^{n+1}}{n+1}} \\ \ln{ \left( 1 - z \right) } &= -\sum_{n = 0}^{\infty}{ \frac{z^{n+1}}{n+1} } \end{align*}$
which is convergent where $\displaystyle \begin{align*} -1 \leq z < 1 \end{align*}$. Don't believe me? When $\displaystyle \begin{align*} z = -1 \end{align*}$ we have $\displaystyle \begin{align*} \ln{ \left( 2 \right) } = \ln{ \left[ 1 - \left( -1 \right) \right] } &= -\sum_{n = 0}^{\infty}{\frac{\left( -1 \right)^{n + 1}}{n + 1}} \end{align*}$, which is an alternating series, and since $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{1}{x + 1} \right) = -\frac{1}{\left( x + 1 \right) ^2} \end{align*}$, that means the non-alternating part $\displaystyle \begin{align*} \frac{1}{n + 1} \end{align*}$ is decreasing, and thus is convergent by the Alternating Series Test.So to answer your previous question - in general any series obtained from differentiating or integrating has the same RADIUS of convergence, but may have different convergence behaviour at the ENDPOINTS of that radius.