Uniform convergence of the series on unbounded domain 

1 $$\sum_{n=1}^\infty (-1)^n \frac{x^2 +  n}{n^2} $$
    Is the series converges uniformly $\mathbb R$


I have tired by this result


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*if $\{f_n(x)\}$ is a sequence of a function defined on a domain $D$ such that 


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*$f_n(x) \geq 0$ for all $x \in D$ and for all $n \in \mathbb N$ 

*$f_{n+1}(x) \leq f_n(x)$ for all $x \in D$

*$\sup_{x\in D} \{f_n(x)\} \to 0$ as $n \to \infty$. Then
$\sum_{n=1}^\infty (-1)^{(n+1)}f_n(x)$ converges uniformly on $D$
first two condition this series is satisfied but third is not satisfied, so this is not work



    
*$$ \sum_{n=1}^\infty \frac{x \sin \sqrt{\frac{x}n}}{x +n} $$ Is the series converges uniformly on $[1, +\infty)$ 
    
  

I have tried Abel test and Dirichlet test,but not getting any solution.



    
*Study the uniform convergence of the series on $\mathbb R$ $$ \sum_{n=1}^\infty \frac{x\sin(n^2x)}{n^2} $$
    
  

Suppose this series converges uniformlybto $f$, so for a given $\epsilon > 0$, there exist $N \in \mathbb N$ such that $\left|\sum_{n=1}^k f_n(x) - f(x)\right| < \epsilon$ for all $k \geq N$
Please tell me how to proceed further. Any help would be appreciated , Thank you

For the second problem, we can use the inequality
$$\sin\left(\sqrt{\frac{x}{n}}\right)\ge \sqrt{\frac{x}{x+n}}$$
for $0\le \sqrt{x/n}\le \pi/2$.  
Then, we have
$$\begin{align}
\sum_{n=N}^\infty \frac{x\,\sin\left(\sqrt{\frac{x}{n}}\right)}{x+n}&\ge  \sum_{n=N}^\infty \left(\frac{x}{x+n}\right)^{3/2}\\\\
&\ge \int_N^\infty \left(\frac{x}{x+y}\right)^{3/2}\,dy\\\\
&=\frac{2x^{3/2}}{\sqrt{x+N}}\\\\
&>1
\end{align}$$
whenever $x=N$ for $N\ge1$.  
Therefore, there exists a number $\epsilon>0$ (here $\epsilon=1$ is suitable) so that for all $N'$ there exists an $x\in[1,\infty)$ (here $x=N$), and there exists a number $N>N'$ (here, take any $N>N'\ge 1$) such that $\left|\sum_{n=N}^\infty \frac{x\,\sin\left(\sqrt{\frac{x}{n}}\right)}{x+n}\right|\ge \epsilon$.  
This is the statement of negation of uniform convergence and therefore the series fails to converge uniformly.
