Assume we are given a sequence of power series $$ B_n(x) := \sum_{k\ge 1}b_{n,k}x^k, \qquad n\in\mathbb{N} $$ with real-valued coefficients $(b_{n,k})_{n,k\ge 1}$. Further we are given a power series $$ B(x) := \sum_{k\ge 1}b_k x^k $$ with real-valued coefficients $(b_k)_{k\ge 1}$ which converges absolutely within its radius of convergence $\rho>0$. Suppose that $B_n(x)$ converges uniformly to $B(x)$ on some interval $[0,a]$ where $a<\rho$.
I want to show that for a given $\varepsilon>0$ there exists $K_0\in\mathbb{N}$ such that uniformly in $x\in[0,a]$ $$ \label{test} \left\lvert \sum_{k >K} b_{n,k}x^k\right\rvert < \varepsilon, \qquad K\ge K_0. \tag{1} $$ I know that we can find such a $K_0$ for $(b_k)_{k\ge 1}$ because absolute convergence of $B(x)$ implies for all $x\in [0,a]$ that \begin{equation} \left\lvert \sum_{k >K} b_{k}x^k\right\rvert \le \sum_{k >K}\lvert b_k\rvert a^k \tag{2}\end{equation} gets arbitrary small for large $K$. But a simple example shows that the coefficients of $B_n(x)$ do not necessarily converge to the coefficients of $B(x)$. For instance, consider $B_n(x)=\sin(nx)/n$ and $B(x)=0$ implying that $b_{n,1}=1$ for all $n\in\mathbb{N}$ and $b_1=0$ (I found this example here: https://math.stackexchange.com/a/2774090/406139).
I believe that $(1)$ is true because of $(2)$ and the uniform convergence of $B_n(x)$ to $B(x)$ but I fail to show this. Does anyone have an idea how to tackle this problem? Thank you :-)
Edit: After some detailed answers to this question, I realised that $(1)$ does not hold in general under the conditions stated above. Again considering $$B_n(x) = \frac{sin(nx)}{n} = \frac{1}{n}\sum_{k\ge 0}(-1)^k \frac{(nx)^{2k+1}}{(2k+1)!} $$ we have for any $x>0$ and $K\in\mathbb{N}$ $$ \frac{1}{n}\sum_{0\le k\le K} (-1)^k \frac{(nx)^{2k+1}}{(2k+1)!} =\Theta\left( n^{2K}\right) $$ which tends to infinty as $n\to\infty$. Hence the sum starting from $K$ also needs to tend to infinity as $B_n(x)$ is bounded for all $n\in\mathbb{N}$.. I guess imposing something like $B_n(x)$ having only non-negative coefficients or converging absolutely is needed.
Unfortunately, coefficients don't necessarily converge in the real case. I don't know about the complex case, but certainly the real case (about which you are enquiring) doesn't work.
The important tool here is the Stone-Weierstrass theorem. This theorem states that the polynomials are dense in the space of continuous real functions on a compact interval, when equipped with the $\infty$-norm (i.e. dense under uniform convergence). That is, given any compact interval $I$, continuous function $f$, and $\varepsilon > 0$, we can find some polynomial $p$ such that $$\|f - p\|_\infty := \sup_{x \in I}|f(x) - p(x)| < \varepsilon.$$
How do we use this? You take a sequence of functions, similar to the one in your question, that converges uniformly to $0$, and approximate this by a polynomial. Let's take $$f_n(x) = n^{-1}\sin(n^2x).$$ Clearly $f_n \to 0$ uniformly. Let's choose $I = [-2, 2]$, and note that this is (well) within the interval of convergence of the $0$ power series, and any other polynomials.
Consider the derivative $$f'_n(x) = 2n\cos(n^2x) - n^{-2}\sin(n^2x).$$ By the Stone-Weierstrass theorem, there exists some polynomial $q_n$ such that $\|f'_n - q_n\| < \frac{1}{2n}$ on $I$. Let $$p_n(x) = \int_0^x q_n(t) \, \mathrm{d}t.$$ Note that $p_n$ is a polynomial with $p_n(0) = 0$ and $p'_n(x) = q_n(x)$. Further, by the fundamental theorem of calculus, given $f_n(0) = 0$, if $x \ge 0$, then \begin{align*} |p_n(x) - f_n(x)| &= \left| \int_0^x (q_n(x) - f'_n(x))\, \mathrm{d}x\right| \\ &\le \int_0^x |q_n(x) - f'_n(x)| \, \mathrm{d}x \\ &\le \int_0^x \frac{1}{2n} \, \mathrm{d}x \\ &= \frac{x}{2n} < \frac{2}{2n} = \frac{1}{n}. \end{align*} Similarly, if $x < 0$, then $$|p_n(x) - f_n(x)| = \left| \int_0^x (q_n(x) - f'_n(x))\, \mathrm{d}x\right| = \left| \int_x^0 (q_n(x) - f'_n(x))\, \mathrm{d}x\right|,$$ and we get the same result similarly. That is, $$\|p_n - f_n\|_\infty \le \frac{1}{n},$$ and so $$\|p_n\|_\infty \le \|p_n - f_n\|_\infty + \|f_n\|_\infty \le \frac{2}{n}.$$ Hence, $p_n \to 0$, uniformly.
Now, we can show that the coefficient of $x$ in $p_n$ will not converge to $0$. Note that this coefficient is just $q_n(0)$, as $q_n(0) = p_n'(0)$. We have that $$\frac{1}{2n} > \|q_n - f'_n\|_\infty \ge |q_n(0) - f'_n(0)| = |q_n(0) - 2n|.$$ Thus, $$q_n(0) > 2n - \frac{1}{2n} \to \infty$$ as $n \to \infty$. So, the coefficient of $x$ need not converge, let alone converge to the correct value, when power series converge uniformly to even a nice function like the $0$ function.