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


*

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


*

*$f_n(x) \geq 0$ 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
 A: *

*In general, if $\sum f_n(x)$ converges uniformly on $\mathbb R,$ then


$$ \sup_{\mathbb R}|f_n| \to 0 \text { as } n\to \infty.$$
This fails in your problem, because in this case $\sup |f_n| \ge |f_n(n^2)| = 1+1/n \not \to 0.$


*We can use the same idea as in 1. In this problem, $f_n(n) = (1/2)\sin 1 \not \to 0,$ so the series doesn't converge uniformly on $[1,\infty).$

*The series converges uniformly on any $[-a,a].$ Proof: We use Weierstrass M:
$$\sup_{[-a,a]}|f_n| \le \frac{a\cdot 1}{n^2}.$$
Since $\sum a/n^2 < \infty$ we have $\sum f_n$ converging uniformly on $[-a,a].$
However, the series does not converge uniformly on $\mathbb R.$ We can again use the idea in 1. Let $x_n = \pi/2n^2 + 2\pi n^2.$ Then verify
$$\sup_{\mathbb R} |f_n| \ge |f_n(x_n)|\ge 1 \not \to 0.$$ 
A: Regarding the first series $$\sum_{n=1}^\infty (-1)^n \frac{x^2 +  n}{n^2} $$ the remainder $$R_{2n}(x) = \sum_{k=2n}^\infty (-1)^n \frac{x^2 +  k}{k^2}$$ can be evaluated grouping one even term with one odd term. You get $$R_{2n}(x) = x^2\sum_{k=2n}^\infty \frac{2k+1}{k^2(k+1)^2} + \sum_{k=2n}^\infty \frac{1}{k(k+1)}$$ The second term of the RHS is converging to $0$ as the series $\sum \frac {1}{k(k+1)}$ is convergent.
Regarding the first one, we have $\frac{2k+1}{k^2(k+1)^2} \sim \frac{2}{k^3}$. Hence $$\sum_{k=2n}^\infty \frac{2k+1}{k^2(k+1)^2} \sim \frac{A}{n^2}$$ with $A > 0$. Therefore $R_{2n}(x)$ doesn't converge uniformly to zero (consider $x=n$). And the series is not uniformly convergent.
A: 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.
