Show that $\phi(x):=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}(1+\frac{x^{2}}{n})^{n}}$ is differentiable on $\mathbb{R}$. Consider the alternating series $$\phi(x):=\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{\sqrt{n}(1+\frac{x^{2}}{n})^{n}}.$$ The exercise asks me to show that this series does not converge absolutely on $\mathbb{R}$, converges uniform on $\mathbb{R}$ and $\phi(x)$ is differentiable on $\mathbb{R}$.
I have proved the first two, but got stuck in the last one. To show $\phi(x)$ is differentiable on $\mathbb{R}$, we need to show that $$(1)\ \ u_{n}(x):=\dfrac{(-1)^{n}}{\sqrt{n}(1+\frac{u^{2}}{n})^{n}}\ \ \text{is differentiable on}\ \ [-R,R];$$ $$(2)\ \ \sum_{n=1}^{\infty}u_{n}'\ \ \text{converges uniformly to some}\ \ g\ \ \text{on}\ \ [-R,R];$$ $$(3)\ \ \sum_{n=1}^{\infty}u_{n}(x_{0})\longrightarrow\phi(x_{0})\ \ \text{for some}\ \ x_{0}\in[ -R,R],$$ where $R$ is arbitrarily fixed. Then, $\phi$ is differentiable on $[-R,R]$ and $\phi'=g$. But $R$ is arbitrary, so taking $R\rightarrow\infty$ finishes the proof.
$(1)$ is clear and the derivative is $$u_{n}'(x)=\dfrac{(-1)^{n+1}2x}{\sqrt{n}(1+\frac{x^{2}}{n})^{n+1}},\ \ \text{for}\ \ n\geq 1.$$
$(3)$ is also clear. Since $(1+\frac{x^{2}}{n})^{n}\nearrow e^{x^{2}}$ and $\sqrt{n}\nearrow \infty$, we have $\frac{1}{\sqrt{n}(1+\frac{x^{2}}{n})^{n}}\searrow 0.$ Hence the alternating series converges pointwise on the whole $\mathbb{R}$ by the alternating series test.
However, I don't know how to show $(2)$. I want to use the following fact:

If a alternating series passes the alternating series test, then we must have $$\Big|\sum_{k=n}^{\infty}a_{k}\Big|\leq |a_{n}|.$$

I used this fact to prove the uniform convergence. However, the denominator of $u_{n}'$ does not have the same monotonicity:
$\sqrt{n}\nearrow\infty$, but $(1+\frac{x^{2}}{n})^{n+1}\searrow e^{x^{2}}$. Thus, even though on $[-R,R]$, $$\frac{2x}{\sqrt{n}(1+\frac{x^{2}}{n})^{n+1}}\rightarrow 0,$$ it is unknown that if this convergence is monotonically decreasing, but this is required by the alternating series test.
Is there any other way to prove the uniform convergence of $\sum u_{n}'$? or I am missing something?
Thank you!
Edit 1:
Below is how I proved the uniform convergence of $\sum_{n=1}^{\infty}u_{n}(x)$.
Let $\epsilon>0$, take $N:=\frac{1}{\epsilon^{2}}-1$, then as the alternating series passes the alternating series test, we use the referred fact above, and see that for all $n\geq N$, and for all $x\in\mathbb{R}$, we have
\begin{align*}
\Big|\sum_{k=n+1}^{\infty}u_{k}(x)\Big|\leq |u_{n+1}(x)|&=\Big|\dfrac{(-1)^{n+1}}{\sqrt{n+1}(1+\frac{x^{2}}{n+1})^{n+1}}\Big|\\
&=\dfrac{1}{\sqrt{n+1}}\Big|\dfrac{1}{(1+\frac{x^{2}}{n+1})^{n+1}}\Big|\\
&\leq\dfrac{1}{\sqrt{n+1}}\\
&\leq \dfrac{1}{\sqrt{N+1}}\\
&=\dfrac{1}{\sqrt{\frac{1}{\epsilon^{2}}-1+1}}\\
&=\epsilon.
\end{align*}
As I argued before, this proof requires the alternating series to pass the alternating series test, but the coefficient of $u_{n}'(x)$ may not decrease to $0$. (It indeed goes to $0$).
 A: Note that $2x/\sqrt{n}$ is monotonically decreasing with respect to $n$  and uniformly convergent to $0$ on $[0,R]$, We also have $-2x/\sqrt{n}$ montonically decreasing with respect to $n$ and uniformly convergent to $0$ on $[-R,0)$. Hence, the series
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}2x}{\sqrt{n}}$$
converges uniformly by Dirichlet's test on both $[-R,0)$ and $[0,R]$, and, therefore, on $[-R,R]$.
The sequence $(1 + x^2/n)^{-(n+1)}$ is eventually monotone  and uniformly bounded for $x \in [-R,R]$. Therefore, by Abel's test we have uniform convergence of
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}2x}{\sqrt{n}} \cdot \frac{1}{\left(1 + \frac{x^2}{n} \right)^{n+1}}$$
A: Knowing that $\sum_{n=1}^{\infty} u_n(x)$ converges locally uniformly, here is a quick proof: Write
\begin{align*}
u_n'(x)
= -\frac{2x}{1+\frac{x^2}{n}} u_n(x)
= -2x u_n(x) + \frac{2x^3}{n+x^2} u_n(x).
\end{align*}
Then on each interval $[-R, R]$,

*

*$\sum_{n=1}^{\infty} (-2x) u_n(x)$ converges uniformly, and


*$\left| \frac{2x^3}{n+x^2} u_n(x) \right| \leq \frac{2R^3}{n^{3/2}} $ uniformly in $n$ and $x$, and so, $\sum_{n=1}^{\infty} \frac{2x^3}{n+x^2} u_n(x)$ converges uniformly by the Weierstrass M-test.
Therefore $\sum_{n=1}^{\infty} u_n'(x)$ also converges uniformly on $[-R, R]$.

Addendum. Another purpose of this answer is to hint a much more general idea: You may split the sum into two parts: conditionally convergent part (which is easier to control) and absolutely convergent part.
In OP's case, we easily check that, for each $R> 0$,
$$ \frac{1}{(1+\frac{x^2}{n})^{n+1}} = e^{-x^2} + \mathcal{O}\left(\frac{1}{n}\right) $$
uniformly in $n$ and $x \in [-R, R]$, where the implicit bound of $\mathcal{O}(\frac{1}{n})$ depends only on $R$. Using this, we may write
$$ u_n'(x) = \frac{(-1)^{n+1}2xe^{-x^2}}{\sqrt{n}} + \mathcal{O}\left(\frac{1}{n^{3/2}}\right), $$
which can be used to easily prove the uniform convergence of $\sum_{n=1}^{\infty} u_n'(x)$ over $[-R, R]$.
