Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$ I am trying to show the followng:
Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$.
I have shown it for a compact subset of $\mathbb{R}$ however, do not know how to extend it to the reals. Below is my proof for convergence on a compact subset.
I break up the sum into two parts and attack each individually 
$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ = 
$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1)}_{(1)}$ + 
$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}[\sin(1 + \frac{x}{n})-\sin(1)]}_{(2)}$.
$\textbf{Equation (1)}$
$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1)$ converges because $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}$ converges by the alternating series test.
$\textbf{Equation (2)}$
we will first bound $\sin(1 + \frac{x}{n})-\sin(1)$ as follows:
\begin{align*}
|\sin(1 + \frac{x}{n})-\sin(1)| &= \Big|\int_{1}^{1+\frac{x}{n}} \cos(t)dt\Big| \\
                                &\leq \Big|\int_{1}^{1+\frac{x}{n}} |\cos(t)| dt\Big| \\
                                &\leq \Big|\int_{1}^{1+\frac{x}{n}} dt\Big| \\
                                &\leq \Big|1+\frac{x}{n} - 1 \Big| \\
                                &= \frac{|x|}{n} \\
\end{align*}
Therefore, on any compact interval $ x \in [-M, M]$
\begin{align*}
|\sin(1 + \frac{x}{n})-\sin(1)| \leq \frac{M}{n}
\end{align*}
It follows that 
\begin{align*}
\Big|\frac{(-1)^{n}}{\sqrt{n}} \Big( \sin(1 + \frac{x}{n})-\sin(1)\Big)\Big| &= \Big|\frac{(-1)^{n}}{\sqrt{n}}\Big|\Big|\sin(1 + \frac{x}{n})-\sin(1)\Big| \\
&= \frac{1}{\sqrt{n}}\Big|\sin(1 + \frac{x}{n})-\sin(1)\Big| \\
&\leq \frac{1}{\sqrt{n}}\frac{M}{n} \\
&= \frac{M}{n^{\frac{3}{2}}}
\end{align*}
Hence,
\begin{align*}
\sum_{n}^{\infty}\frac{1}{n^{\frac{3}{2}}} < \infty &\implies \sum_{n}^{\infty}\frac{M}{n^{\frac{3}{2}}} < \infty \\
&\implies \sum_{n=1}^{\infty}\Big|\frac{(-1)^{n}}{\sqrt{n}} \Big[ \sin(1 + \frac{x}{n})-\sin(1)\Big]\Big| < \infty \\
&\implies \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}} \Big[ \sin(1 + \frac{x}{n})-\sin(1)\Big] < \infty
\end{align*}
Hence, we have that both $(1)$ and $(2)$ converge on a compact interval $[-M,M]$ which implies our original equation of interest does also.
I want to extend this proof to all $\mathbb{R}$ but I do not know how to. Could someone please help me with this. Anything is appreciated.
 A: The sum is not uniformly convergent on $\mathbb{R}$.
Take $\epsilon \in (0,\frac{1}{10})$. Suppose $N$ is large such that $|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})| \le \epsilon$ for all $x \in \mathbb{R}$. Take $X = 2\pi\prod_{N \le n \le 2N} n$. Independent of the specific choice of $X$, we of course have $$\frac{1}{X}\int_0^{X} \left|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})\right|^2dx \le \epsilon^2.$$ But due to the specific choice, we have $$\frac{1}{X}\int_0^{X} \left|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})\right|^2dx$$ $$= \frac{1}{X}\int_0^X \sum_{N \le n,m \le 2N} \frac{(-1)^{n+m}}{\sqrt{nm}}\sin(1+\frac{x}{n})\sin(1+\frac{x}{m})dx$$ $$ = \frac{1}{X}\sum_{N \le n \le 2N} \frac{1}{n}\left(\frac{X}{2}-\frac{1}{4}n\sin(\frac{2(n+X)}{n})+\frac{1}{4}n\sin(2)\right),$$ where we used $\int \sin(1+\frac{x}{n})\sin(1+\frac{x}{m})dx = \frac{1}{2}mn\left(\frac{\sin(x(\frac{1}{n}-\frac{1}{m}))}{m-n}-\frac{\sin(\frac{x}{m}+\frac{x}{n}+2)}{m+n}\right)$ for $m \not = n$ (which takes the same value at $0$ and $X$ by our choice of $X$) and $\int \sin^2(1+\frac{x}{n})dx = \frac{x}{2}-\frac{1}{4}n\sin(\frac{2(n+x)}{n})$. And once again by our choice of $X$, we end up with $$\frac{1}{X}\sum_{N \le n \le 2N} \frac{X}{2n},$$ which is around $\frac{1}{2}\ln(2)$, much greater than $\frac{1}{100}$.
