Uniform convergence of $\sum\limits_{n=1}^{+\infty} \left(\sin{{1} \over {n}}\right) x^n$ $f_n(x)=\left(\sin{{1} \over {n}}\right) x^n$
pointwise convergence: $|f_n(x)|=\left(\sin{{1} \over {n}}\right) |x|^n \sim {{|x|^n}\over{n}}$ for $n \rightarrow +\infty$
$\sum\limits_{n=1}^{+\infty}{{|x|^n}\over{n}}$ is a power series and it converges in $E=[-1,1)$
uniform convergence: I calculate sup$_E|f_n(x)|=f_n(1)$ so there isn't unif. convergence in E. Can I have convergence in a subset of E?
If I consider interval $E=[a,b], -1<a<b<1$, $\sup_E|f_n(x)|=\max\left\{\sin{{1} \over {n}}|a|^n,\sin{{1} \over {n}}|b|^n\right\}$, general term of convergent series so there is uniform convergence in $[a,b]$?
 A: The series converges uniformly on any compact interval $[-1,a] \subset [-1,1)$ with $-1$ as the left endpoint by the Dirichlet test.
Note that $\sin \frac{1}{n} \searrow 0$ monotonically and $\sum_{n = 1}^m x^n$ is uniformly bounded for all $m \in \mathbb{N}$ and all $x \in [-1,a]$ with
$$\left|\sum_{n = 1}^m x^n \right| \leqslant \max(1,|a|/(1-|a|)$$
However even though the series converges pointwise for each $x \in [-1,1)$ the convergence is not uniform on any interval $(a,1)$.  We can take $a = 0$ with no loss of generality to prove this non-uniform convergence.
Note that 
$$\sup_{x \in (0,1)} \left|\sum_{k=n+1}^{\infty}x^k \sin \frac{1}{k}\right| > \sup_{x \in (0,1)} \sum_{k=n+1}^{2n}x^k \sin \frac{1}{k} \\>  \sup_{x \in (0,1)} nx^{2n}\sin \frac{1}{2n} = \frac{1}{2}\frac{\sin \frac{1}{2n}}{\frac{1}{2n}}$$
Since the RHS converges to $\frac{1}{2}$ as $n \to \infty$ the Cauchy criterion for uniform convergence is violated.
A: Note that the series $\sum_{n=1}^\infty \sin\left(\frac1n\right)x^n$ fails to converge uniformly on $(-1,1)$.  To see this, we can write
$$\sum_{n=1}^\infty \sin\left(\frac1n\right)x^n=\sum_{n=1}^\infty \left(\sin\left(\frac1n\right)-\frac1n\right)x^n+\sum_{n=1}^\infty \frac{x^n}n\tag1$$
The first series on the right-hand side of $(1)$ converges uniformly on $[-1,1]$ since the term $\sin\left(\frac1n\right)-\frac1n=O\left(\frac1{n^3}\right)$.  Therefore, it is enough to show that the second series on the right-hand side of $(1)$ fails to converge on $(-1,1)$.

To negate the uniform convergence of $\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x)$ on $(-1,1)$, we choose $\varepsilon=\frac18\log(2)$.  Then, we have for any $N\ge1$ and $x=1-\frac1{N+1}\in(0,1)$
$$\begin{align}
\left|-\log(1-x)-\sum_{n=1}^N \frac{x^n}{n}\right|=&\left|\sum_{n=1}^\infty \frac{x^n}{n}-\sum_{n=1}^{N} \frac{x^n}{n}\right|\\\\
&=\sum_{n=N+1}^\infty \frac{x^n}{n}\\\\
&\ge \sum_{n=N+1}^{2N+1} \frac{x^n}{n}\\\\
&\ge \left(1-\frac1{N+1}\right)^{2N+1}\sum_{n=N+1}^{2N+1}\frac1n\\\\
&\ge \left(1-\frac1{N+1}\right)^{2N+1}\log(2)\\\\
&\ge \frac18\log(2)\\\\
&=\varepsilon
\end{align}$$ 
And hence the series $\sum_{n=1}^\infty \frac{x^n}{n}$ converges on $[-1,1)$ but fails to converge uniformly on $[-1,1)$.
