Show that $\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n+x^4}$ is uniformly convergent on $\Bbb{R}$ Show that the following series is uniformly convergent on $\Bbb{R}$
\begin{align}\sum^{\infty}_{n=1}(-1)^{n+1}\dfrac{1}{n+x^4}\end{align}
MY TRIAL 
I tried using the alternating series test before the $\beta_n$ approach.
Let $f_n(x)=\dfrac{1}{n+x^4},\;\forall\;x\in\Bbb{R},\;n\in\Bbb{N},$
then 


*

*$f_n(x)=\dfrac{1}{n+x^4}\geq 0$

*$f_{n+1}(x)\leq f_{n}(x)$

*$f_n(x)=\dfrac{1}{n+x^4}\to 0$


Then,
\begin{align}\beta_n &=\sup\limits_{x\in\Bbb{R}}\left|s_n(x)-\sum^{\infty}_{i=1}f_i(x)\right|\\&=\sup\limits_{x\in\Bbb{R}}\left|\sum^{n}_{i=1}(-1)^{i+1}f_i(x)-\sum^{\infty}_{i=1}(-1)^{i+1}f_i(x)\right|\\&=\sup\limits_{x\in\Bbb{R}}\left|\sum^{n}_{i=1}(-1)^{i+1}\dfrac{1}{i+x^4}-\sum^{\infty}_{i=1}(-1)^{i+1}\dfrac{1}{i+x^4}\right|\\&=\sup\limits_{x\in\Bbb{R}}\left|(-1)^{n+2}\dfrac{1}{(n+1)+x^4}+(-1)^{n+3}\dfrac{1}{(n+2)+x^4}+(-1)^{n+4}\dfrac{1}{(n+3)+x^4}\cdots\right|\\&=\sup\limits_{x\in\Bbb{R}}\left|\dfrac{1}{(n+1)+x^4}-\dfrac{1}{(n+2)+x^4}+\dfrac{1}{(n+3)+x^4}-\dfrac{1}{(n+4)+x^4}\cdots\right|\\&=\sup\limits_{x\in\Bbb{R}}\left|\dfrac{1}{(n+1)+x^4}-\left(\dfrac{1}{(n+2)+x^4}-\dfrac{1}{(n+3)+x^4}\right)-\left(\dfrac{1}{(n+4)+x^4}-\dfrac{1}{(n+5)+x^4}\right)\cdots\right|\\&\leq \sup\limits_{x\in\Bbb{R}}\left|\dfrac{1}{(n+1)+x^4}\right|\to 0,\;\;\text{as}\;n\to\infty\end{align}
and we are done!
Kindly help me check if I'm correct! Constructive criticisms will be highly welcome! I'll also love to see other approaches to this problem. Thanks!
 A: For future readers, I present two proofs:
METHOD I: DIRICHLET'S TEST
The Dirichlet's test for uniform convergence can be found here.  
Let $\;\epsilon>0,\; x\in\Bbb{R}$ and $n\in\Bbb{N}$ be fixed but arbitrary. Define
\begin{align}f_n(x)=(-1)^{n+1}\;\;\text{and }\;\;g_{n}(x)=\dfrac{1}{n+x^4}\end{align}
Then, uniform boundedness is implied, since
\begin{align}
  &\left|F_n(x)\right|=\left|\sum^{n}_{m}(-1)^{m+1} \right|\leq 1,\end{align}
The sequence $\{g_n(x)\}_{n\in \Bbb{N}}$ is decreasing, since
\begin{align}
  \dfrac{n+x^4}{(n+1)+x^4}\leq \dfrac{(n+1)+x^4}{(n+1)+x^4}=1\implies g_{n+1}(x)\leq g_{n}(x)\end{align}
By Archmidean principle, there exists $N(\epsilon)$ such that $N(\epsilon)>\epsilon$ and
\begin{align}
  &\left|g_n(x)-0\right|=\left|\dfrac{1}{n+x^4}-0 \right|=\dfrac{1}{n+x^4}\leq \dfrac{1}{n}\leq \dfrac{1}{N}<\epsilon,\;\forall\;n\geq N(\epsilon)\end{align}
Hence, $\{g_n(x)\}_{n\in \Bbb{N}}$ converges uniformly on $\Bbb{R}$ and, by Dirichlet's test, $\sum^{\infty}_{n=1}f_n(x)g_n(x)$ converges uniformly on $\Bbb{R}$.
METHOD II: UNIFORM CAUCHY CRITERION
Let $\;\epsilon>0$, $m,n\in\Bbb{N}$ be given such that $m\geq n$ and $x\in\Bbb{R}$ be arbitrary. Define $\{f_n(x)\}_{n\in \Bbb{N}}$ and $\{g_n(x)\}_{n\in \Bbb{N}}$ as before. Then,
\begin{align}\left|\sum^{m}_{k=1}f_k(x)g_k(x)-\sum^{n}_{k=1}f_k(x)g_k(x) \right|&=\left|\sum^{m}_{k=n+1}f_k(x)g_k(x)\right|\\&=\left|\sum^{m}_{k=n+1}(-1)^{k+1}\dfrac{1}{k+x^4}\right|\end{align}
Expanding, we get
\begin{align}\left|\sum^{m}_{k=n+1}(-1)^{k+1}\dfrac{1}{k+x^4}\right|&=\left|(-1)^{n+2}\dfrac{1}{(n+1)+x^4}+(-1)^{n+3}\dfrac{1}{(n+2)+x^4}+\cdots +(-1)^{n+1}\dfrac{1}{n+x^4}\right|\\&=\left|\dfrac{1}{(n+1)+x^4}-\dfrac{1}{(n+2)+x^4}+\cdots +(-1)^{m-n-2}\dfrac{1}{(m-1)+x^4}+(-1)^{m-n-1}\dfrac{1}{m+x^4}\right|\\&\leq \left|\dfrac{1}{(n+1)+x^4}-\dfrac{1}{(n+2)+x^4}\right|+\cdots +\left|\dfrac{1}{(m-1)+x^4}-\dfrac{1}{m+x^4}\right|\\&= \dfrac{1}{(n+1)+x^4}-\dfrac{1}{(n+2)+x^4}+\cdots +\dfrac{1}{(m-1)+x^4}-\dfrac{1}{m+x^4}\\&= \dfrac{1}{(n+1)+x^4}-\left[\dfrac{1}{(n+2)+x^4}-\dfrac{1}{(n+2)+x^4}\right]-\cdots \\&\;-\left[\dfrac{1}{(m-1)+x^4}-\dfrac{1}{(m-1)+x^4}\right]-\dfrac{1}{m+x^4}\\&\leq \dfrac{1}{(n+1)+x^4}.\end{align}
As shown in Method I, there exists $N(\epsilon)$ such that $,\;\forall\;n\geq N(\epsilon)$,
\begin{align}
  \dfrac{1}{(n+1)+x^4}\leq \dfrac{1}{n+x^4}\leq \dfrac{1}{n}\leq \dfrac{1}{N}<\epsilon,\end{align}
and we're done!
