For a power series $\sum_n c_nx^n$, the radius of convergence is $\sup\{R>0, \{c_nx^n\}\mbox{ is bounded if }|x|\leq R\}$.
Let $x$ such that $|x|<R_1$. Then $|(a_n+b_n)x^n|\leq |a_n||x|^n+|b_n||x^n|$ and the sequence $\{(a_n+b_n)x^n\}$ is bounded. Hence the radius of convergence of $\sum_n(a_n+b_n)x^n$ is $\geq R_1$. If we take $|x|\in (R_1,R_2)$, then $\{(a_n+b_n)x^n\}$ is not bounded, as a sum of a bounded sequence with an unbounded one. So the radius of convergence is at most $R_1$.
Conclusion: the radius of convergence of $\sum_n (a_n+b_n)x^n$ is $R_1$.
Now we look at the case $R_1=R_2$. Let $R>R_1$ and $\sum_nc_nx^n$ a series or radius of convergence $R$. By the previous case, $\sum_n(-a_n+c_n)x^n$ has a radius of convergence $R_1$ but $\sum_n(a_n+(-a_n+c_n))x^n$ has a radius of convergence $R$. So the sum of two series of radius of convergence $R_1$ can be of radius of convergence $R\geq R_1$.
Suppose that a sequence $f_n(x)$ converges pointwise to the function $f(x)$ for all $x\in S$.
NEGATION OF UNIFORM CONVERGENCE
The sequence $f_n(x)$ fails to converge uniformly to $f(x)$ for $x\in S$ if there exists a number $\epsilon>0$ such that for all $N$, there exists an $n_0>N$ and a number $x\in S$ such that $|f_{n_0}(x)-f(x)|\ge \epsilon$.
Now, let $f_n(x)=a_nx^n$ with $\lim_{n\to \infty}f_n(x)=0$ for all $x\in \mathbb{R}$. Certainly, either $a_n=0$ for all $n$ sufficiently large or for any number $N$ there exists a number $n_0>N$ such that $a_{n_0}\ne 0$.
Suppose that the latter case holds. Now, taking $\epsilon=1$, we find that
$$|a_{n_0}x^{n_0}|\ge \epsilon$$
whenever $|x|\ge |a_{n_0}|^{-1/n_0}$. And this negates the uniform convergence of $f_n(x)$.
And inasmuch as the sequence $f_n(x)$ fails to uniformly converge to zero, then the series $\sum_{n=0}^\infty f_n(x)$ fails to uniformly converge.
Note for the example for which $a_n=n^n$ we can take $x>1/n$.
NOTE:
If $a_n=0$ for all $n>N+1$, then we have
$$\sum_{n=0}^\infty a_nx^n = \sum_{n=0}^N a_nx^n$$
which is a finite sum and there is no issue regarding convergence.
Best Answer
Since the radius of convergence of $\sum_{n=0}^\infty a_nx^n$ is greater than $0$, $\limsup_n\sqrt[n]{|a_n}<\infty$. Take $r>0$ such that $\limsup_n\sqrt[n]{|a_n|}<r$. Then, if $n\gg0$, $\sqrt[n]{|a_n|}<r$. Since the inequality $\sqrt[n]{|a_n|}\geqslant r$ can take place only for finitely many $n$'s, if you replace $r$ by some larger number $R$, you will have $(\forall n\in\Bbb N):\sqrt[n]{|a_n|}<R$. But then $|a_n|<R^n$, and so$$|b_n|=\left|\sum_{k=0}^na_k\right|\leqslant1+R+R^2+\cdots+R^n=\frac{R^{n+1}-1}{R-1}.$$But then$$\limsup_n\sqrt[n]{|b_n|}\leqslant R.$$