Radius of convergence of a derivative of a power series. The power series $\sum_{0}^{\infty}k_n(x-b)^n$ and $\sum_{1}^{\infty}nk_n(x-b)^{n-1}$ have the same radius of convergence, however would it be true to say that $\sum_{1}^{\infty}k_n(x-b)^n$ and $\sum_{1}^{\infty}nk_n(x-b)^{n-1}$ have the same radii of convergence?
I have to be careful as this is part of a homework question, but i feel as though just knowing this fact is not cheating. 
In my example, i have 'Taken out' the first term in the second sum to give n = 2 in the lower limit of the sum. This fits the 'rule', However i now have this;
$\sum_{1}^{\infty}nk_n(x-b)^{n-1} = 1/t + \sum_{2}^{\infty}nk_n(x-b)^{n-1}$ where $t$ is some integer. 
Surely this must affect on the radius on convergence. Could someone clarify that it does/does not.
I seem to think the new radius of convergence is $R + 1/t$

No, it does not affect the radius of convergence. The convergence of a series does not change if a finite number of terms are omitted. That is
$$
\sum_{n=0}^\infty a_n\text{ converges if and only if }\sum_{n=N}^\infty a_n\text{ converges,}
$$
where $N$ is a positive integer.
