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!

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!
